CN109612870B - Method for conveniently measuring total organic matter and moisture content of plant vinegar - Google Patents
Method for conveniently measuring total organic matter and moisture content of plant vinegar Download PDFInfo
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- 235000021419 vinegar Nutrition 0.000 title claims abstract description 174
- 239000000052 vinegar Substances 0.000 title claims abstract description 174
- 239000005416 organic matter Substances 0.000 title claims abstract description 124
- 238000000034 method Methods 0.000 title claims abstract description 81
- XLYOFNOQVPJJNP-UHFFFAOYSA-N water Substances O XLYOFNOQVPJJNP-UHFFFAOYSA-N 0.000 claims abstract description 151
- 239000007788 liquid Substances 0.000 claims abstract description 101
- 238000010438 heat treatment Methods 0.000 claims abstract description 42
- 238000005070 sampling Methods 0.000 claims abstract description 34
- 239000007864 aqueous solution Substances 0.000 claims abstract description 14
- 239000000126 substance Substances 0.000 claims abstract description 6
- 238000005303 weighing Methods 0.000 claims abstract description 5
- 150000007529 inorganic bases Chemical class 0.000 claims description 121
- HEMHJVSKTPXQMS-UHFFFAOYSA-M Sodium hydroxide Chemical compound [OH-].[Na+] HEMHJVSKTPXQMS-UHFFFAOYSA-M 0.000 claims description 120
- 239000012153 distilled water Substances 0.000 claims description 80
- 241000196324 Embryophyta Species 0.000 claims description 59
- 238000007865 diluting Methods 0.000 claims description 54
- 235000017166 Bambusa arundinacea Nutrition 0.000 claims description 52
- 235000017491 Bambusa tulda Nutrition 0.000 claims description 52
- 241001330002 Bambuseae Species 0.000 claims description 52
- 235000015334 Phyllostachys viridis Nutrition 0.000 claims description 52
- 239000011425 bamboo Substances 0.000 claims description 52
- 238000001035 drying Methods 0.000 claims description 40
- 238000004448 titration Methods 0.000 claims description 39
- 244000025254 Cannabis sativa Species 0.000 claims description 31
- 239000000243 solution Substances 0.000 claims description 21
- 238000010790 dilution Methods 0.000 claims description 20
- 239000012895 dilution Substances 0.000 claims description 20
- 239000002023 wood Substances 0.000 claims description 5
- 238000004364 calculation method Methods 0.000 claims description 4
- LUMVCLJFHCTMCV-UHFFFAOYSA-M potassium;hydroxide;hydrate Chemical compound O.[OH-].[K+] LUMVCLJFHCTMCV-UHFFFAOYSA-M 0.000 claims description 4
- JBJWASZNUJCEKT-UHFFFAOYSA-M sodium;hydroxide;hydrate Chemical compound O.[OH-].[Na+] JBJWASZNUJCEKT-UHFFFAOYSA-M 0.000 claims description 4
- 238000013441 quality evaluation Methods 0.000 abstract description 2
- 239000003513 alkali Substances 0.000 abstract 1
- 239000000463 material Substances 0.000 description 36
- 238000000691 measurement method Methods 0.000 description 13
- 230000000694 effects Effects 0.000 description 4
- OKTJSMMVPCPJKN-UHFFFAOYSA-N Carbon Chemical compound [C] OKTJSMMVPCPJKN-UHFFFAOYSA-N 0.000 description 3
- 238000011161 development Methods 0.000 description 3
- 238000005259 measurement Methods 0.000 description 3
- 239000003153 chemical reaction reagent Substances 0.000 description 2
- 238000004587 chromatography analysis Methods 0.000 description 2
- 238000004108 freeze drying Methods 0.000 description 2
- 239000003960 organic solvent Substances 0.000 description 2
- KMUONIBRACKNSN-UHFFFAOYSA-N potassium dichromate Chemical compound [K+].[K+].[O-][Cr](=O)(=O)O[Cr]([O-])(=O)=O KMUONIBRACKNSN-UHFFFAOYSA-N 0.000 description 2
- 238000000638 solvent extraction Methods 0.000 description 2
- 238000012360 testing method Methods 0.000 description 2
- 238000003109 Karl Fischer titration Methods 0.000 description 1
- 238000010521 absorption reaction Methods 0.000 description 1
- 229910052799 carbon Inorganic materials 0.000 description 1
- 239000003610 charcoal Substances 0.000 description 1
- 238000006243 chemical reaction Methods 0.000 description 1
- 239000003638 chemical reducing agent Substances 0.000 description 1
- 238000003869 coulometry Methods 0.000 description 1
- 238000013461 design Methods 0.000 description 1
- 230000007613 environmental effect Effects 0.000 description 1
- 239000003337 fertilizer Substances 0.000 description 1
- 150000002484 inorganic compounds Chemical class 0.000 description 1
- 229910010272 inorganic material Inorganic materials 0.000 description 1
- 239000011344 liquid material Substances 0.000 description 1
- 150000002894 organic compounds Chemical class 0.000 description 1
- 230000003647 oxidation Effects 0.000 description 1
- 238000007254 oxidation reaction Methods 0.000 description 1
- 239000002689 soil Substances 0.000 description 1
- 239000012086 standard solution Substances 0.000 description 1
- 239000010902 straw Substances 0.000 description 1
- 238000011179 visual inspection Methods 0.000 description 1
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- G—PHYSICS
- G01—MEASURING; TESTING
- G01N—INVESTIGATING OR ANALYSING MATERIALS BY DETERMINING THEIR CHEMICAL OR PHYSICAL PROPERTIES
- G01N5/00—Analysing materials by weighing, e.g. weighing small particles separated from a gas or liquid
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- G—PHYSICS
- G01—MEASURING; TESTING
- G01N—INVESTIGATING OR ANALYSING MATERIALS BY DETERMINING THEIR CHEMICAL OR PHYSICAL PROPERTIES
- G01N5/00—Analysing materials by weighing, e.g. weighing small particles separated from a gas or liquid
- G01N5/04—Analysing materials by weighing, e.g. weighing small particles separated from a gas or liquid by removing a component, e.g. by evaporation, and weighing the remainder
- G01N5/045—Analysing materials by weighing, e.g. weighing small particles separated from a gas or liquid by removing a component, e.g. by evaporation, and weighing the remainder for determining moisture content
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- General Health & Medical Sciences (AREA)
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Abstract
A method for conveniently measuring the total organic matter and water content of plant vinegar belongs to the technical field of natural quality evaluation. The method comprises the following steps: 1) sampling the plant vinegar liquid; 2) titrating the plant vinegar liquid with an inorganic alkali aqueous solution until the alkaline pH value is 7-14; 3) heating the treated plant vinegar liquid to completely volatilize the liquid; 4) weighing the total mass of the remaining substances; 5) calculating the total organic matter percentage content of 6 samples; 6) and calculating the percentage content of the total organic matters and the percentage content of water in the plant vinegar liquid. The method for conveniently measuring the total organic matter and moisture content of the plant vinegar solves the technical problem of measuring the total organic matter and moisture percentage content, and can be used for measuring the total organic matter and moisture percentage content of the plant vinegar in a factory or a laboratory; and is used for designing and producing instruments and equipment related to the percentage content of total organic matters and water in the plant vinegar.
Description
Technical Field
The invention belongs to the technical field of natural quality evaluation, and particularly relates to a method for conveniently measuring the total organic matter and moisture content of field planting vinegar.
Background
Bamboo, wood and grass carbon, especially activated carbon, are high in price and have good profit and market development space, so that the industry of biochar is getting bigger, but the application development of plant vinegar is very slow, and the sustainable development of bamboo charcoal is greatly restricted along with the stricter environmental check. The method is particularly important for measuring the organic matter content of the plant vinegar liquid because the quality of the plant vinegar liquid cannot be evaluated due to the difficulty in measuring the organic matter content proportion of the plant vinegar liquid, so that the sale, popularization and application of the plant vinegar liquid are influenced. In the methods for analyzing the components of the plant vinegar liquid by chromatography, the most important sample treatment method is to remove a large amount of water in the plant vinegar liquid, and some literature reports exist, wherein freeze drying methods, organic solvent extraction methods and the like are mainly adopted, wherein the freeze drying methods are expensive and have no market industrial application prospect, the organic solvent extraction methods are too expensive and are not suitable for conventional determination, and the chromatography combination can only determine the relative proportion of each organic matter. The absolute percentage content of the total organic matter in the plant vinegar liquid cannot be measured. If the method is used for measuring the organic matter content in organic fertilizers and soil, namely a potassium dichromate volumetric method, the method needs strong oxidation reducing agents and unsafe use, and also needs standard solutions and an o-coffee indicator, the measuring method is complicated, hundreds of components of bamboo vinegar greatly interfere with experimental measurement, and the percentage content of organic matters is difficult to accurately measure. If a moisture tester is used for measuring the moisture content of the plant vinegar liquid, the percentage content of the organic matters is obtained by subtracting 100 percent, and the moisture measurement mainly comprises two methods, namely a titration method and a coulometry method by a Karl Fischer method. The method is suitable for measuring the water content in many inorganic compounds and organic compounds. A method for measuring moisture by volumetric analysis is an eye measurement method in GB6283 determination of moisture content in chemical products. Visual inspection can only measure the moisture content of colorless liquid material. The plant vinegar liquid is a dark brown liquid which is not suitable for the method, and then the method is developed into an electric method. However, since the karl fischer titration reagent used easily absorbs moisture, a well-sealed system is required for a burette, a titration cell (measurement cell), and the like of the titrant delivery system. Otherwise, the moisture absorption phenomenon causes long-time instability and serious errors of the end point, and in addition, the components of the bamboo vinegar liquid are complex and diverse, and the reaction on the reagent influences the test result.
Disclosure of Invention
Aiming at the problems in the prior art, the invention aims to design and provide a technical scheme of a method for conveniently measuring the total organic matter and the moisture content of the planting vinegar, solves the technical problem of measuring the total organic matter and the moisture percentage content, and can be used for measuring the total organic matter and the moisture percentage content of the planting vinegar in a factory and a laboratory; and is used for designing and producing instruments and equipment related to the percentage content of total organic matters and water in the plant vinegar.
The method for conveniently measuring the total organic matter and water content of the plant vinegar is characterized by comprising the following steps of:
1) sampling 6-60 g of the plant vinegar, calculating a g, diluting 6 samples by using distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times, wherein the dilution to 1 time means a g of plant vinegar and 0 g of distilled water; diluting to 2 times refers to 1/2 × a g of plant vinegar liquid and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of plant vinegar liquid and 2/3 × a g of distilled water, and so on, and taking a g of samples for later use;
2) titrating the plant vinegar liquid diluted in the step 1) with 5-50% of inorganic base aqueous solution until the alkaline pH value is 7-14, wherein b% of inorganic base aqueous solution is used, and C g of inorganic base aqueous solution is consumed when the titration is finished; the inorganic alkaline water solution is NaOH water solution or KOH water solution;
3) heating the plant vinegar liquid treated in the step 2) to 100-150 ℃ or drying by hot air to completely volatilize the liquid;
4) weighing the total mass of the rest substances, and calculating d grams;
5) the percentage content e% of total organic matter of 6 samples is calculated, and the calculation method comprises the following steps: using NaOH: e% = (d-0.575 bc%)/a × 100%; using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1%, e2%, e3%, e4%, e5%, e 6%;
6) calculating the percentage content e% of total organic matters and the percentage content g% of water of the plant vinegar liquid, taking 1, 2, 3, 4, 5 and 6 as an X axis, taking e1%, e2%, e3%, e4%, e5% and e6% as a Y axis, establishing a regression equation, enabling X =1, and calculating a Y value to be an e% value; the percentage content of water in the plant vinegar liquid is g percent, g percent is not less than 1 percent, and the total organic matter percentage content of the plant vinegar liquid is e percent;
the plant vinegar is one of bamboo vinegar, wood vinegar or grass vinegar.
The method for conveniently measuring the total organic matter and water content of the plant vinegar is characterized by comprising the following steps of:
1) sampling 6-60 g of plant vinegar liquid, and counting as a g for later use;
2) titrating the plant vinegar liquid obtained in the step 1) with 5-50% of inorganic base aqueous solution until the alkaline pH value is 7-14, wherein the b% of inorganic base aqueous solution is used, and C g of inorganic base aqueous solution is consumed when the titration is finished; the inorganic alkaline water solution is NaOH water solution or KOH water solution;
3) heating the plant vinegar liquid treated in the step 2) to the temperature of 100-110 ℃ or drying by hot air to completely volatilize the liquid;
4) weighing the total mass of the rest substances, and calculating d grams;
5) the percentage content e% of total organic matter and the percentage content g% of water of the plant vinegar liquid are calculated, and the calculation method comprises the following steps: using NaOH: e% = (d-0.575 bc%)/a × 100%; using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the plant vinegar liquid is e percent; the percentage content of water in the plant vinegar liquid is g percent, g percent is not less than 1 percent, and the total organic matter percentage content of the plant vinegar liquid is e percent;
the plant vinegar is one of bamboo vinegar, wood vinegar or grass vinegar.
The method for conveniently measuring the total organic matter and water content of the plant vinegar is characterized by comprising the following steps of 1): 10-50 g, preferably 20-40 g, more preferably 25-30 g of the plant vinegar liquid is sampled.
The method for conveniently measuring the total organic matter and water content of the plant vinegar is characterized in that in the step 2): the percentage concentration b% of the aqueous solution of inorganic base is 10 to 40%, preferably 20 to 35%, more preferably 25 to 30%; titration is carried out to an alkaline pH of 8 to 13, preferably 9 to 12, more preferably 10 to 11.
The method for conveniently measuring the total organic matter and water content of the plant vinegar is characterized in that in the step 3): the heating temperature is 110-140 ℃, preferably 120-130 ℃.
The method for conveniently measuring the total organic matter and moisture content of the plant vinegar solves the technical problem of measuring the total organic matter and moisture percentage content, and can be used for measuring the total organic matter and moisture percentage content of the plant vinegar in a factory or a laboratory; and is used for designing and producing instruments and equipment related to the percentage content of total organic matters and water in the plant vinegar. The parts referred to in the present document are parts by mass percentages unless otherwise indicated.
Drawings
FIG. 1 is a process flow section of an embodiment of the present invention.
Detailed Description
The present invention is further illustrated by the following examples.
Example 1
1) Sampling 60 g of bamboo vinegar (a =60 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on) taking 60 g of each 6 samples for later use;
2) titration of 50% aqueous inorganic base KOH to alkaline pH12 (using b% =50% aqueous inorganic base, titration to endpoint consumes C =15.2 grams aqueous inorganic base)
3) Heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =9.9 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (9.9-0.697 × 15.2 × 50%)/60 × 100% =7.9%, similarly e2% =4.0%, e3% =2.7%, e4% =2.0%, e5% =1.7%, e6% =1.4%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 7.9%, 4.0%, 2.7%, 2.0%, 1.7% and 1.4% as a Y axis, establishing a regression equation of Y = -1.1457X + 7.2933 (R = 0.998), enabling X =1, and calculating the Y value as the e% value e% = 6.15%. The bamboo vinegar liquid has the water content (g%) = 1-total organic matter content (e%) =1-6.15% = 93.85%.
Example 2
1) Sampling 30 g of bamboo vinegar (accounting for a =30 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on) taking 30 g of each 6 samples for later use;
2) titration of aqueous NaOH, 50% inorganic base, to alkaline pH11 (b% =50% aqueous inorganic base used, consumption C =7.5 g aqueous inorganic base at endpoint)
3) Heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.7 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (4.7-0.575 × 7.5 × 50%)/30 × 100% =8.5%, similarly e2% =4.2%, e3% =2.8%, e4% =2.1%, e5% =1.8%, e6% =1.5%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 8.5%, 4.2%, 2.8%, 2.1%, 1.8% and 1.5% as a Y axis, establishing a regression equation Y = -1.2257X + 7.7733 (R = 0.869), enabling X =1, and calculating the Y value as the e% value e% = 6.55%; the bamboo vinegar liquid has the water percentage content (g%) = 1-total organic matter percentage content (e%) =1-6.55% = 93.45%.
Example 3
1) Sampling 30 g of bamboo vinegar (accounting for a =30 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on) taking 30 g of each 6 samples for later use;
2) titration of 25% aqueous inorganic base KOH to alkaline pH10 (using b% =25% aqueous inorganic base, titration to endpoint consumes C =14.8 grams aqueous inorganic base)
3) Heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =5.2 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (5.2-0.697 × 14.8 × 25%)/30 × 100% =8.7%, similarly e2% =4.5%, e3% =2.8%, e4% =2.2%, e5% =1.9%, e6% =1.6%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample by taking 1, 2, 3, 4, 5 and 6 as X axes, 8.7%, 4.5%, 2.8%, 2.2%, 1.9% and 1.6% as Y axes, establishing a regression equation of Y = -1.2543X + 8.0067 (R = 0.999), enabling X =1, and calculating the Y value as the e% value e% = 6.75%. The bamboo vinegar liquid has the water percentage content (g%) = 1-total organic matter percentage content (e%) =1-6.75% = 93.25%.
Example 4
1) Sampling 30 g of bamboo vinegar (accounting for a =30 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on) taking 30 g of each 6 samples for later use;
2) aqueous NaOH 20% inorganic base titrated to alkaline PH9 (using aqueous b% =20% inorganic base, consumption C =11.2 g aqueous inorganic base at endpoint)
3) Heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.0 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (4.0-0.575 × 11.2 × 20%)/30 × 100% =9.0%, similarly e2% =4.6%, e3% =2.9%, e4% =2.3%, e5% =1.9%, e6% =1.5%;
6) calculating the percentage content of total organic matters (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 9.0%, 4.6%, 2.9%, 2.3%, 1.9% and 1.5% as a Y axis, establishing a regression equation Y = -1.32X + 8.32 (R = 0.999), making X =1, and calculating the Y value as the e% value e% = 7.0%. The bamboo vinegar liquid has the water content (g%) = 1-total organic matter content (e%) =1-7.0% = 93.0%.
Example 5
1) Sampling 15 g of bamboo vinegar (accounting for a =15 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on), and respectively taking 15 g of 6 samples for later use;
2) titration of aqueous 15% inorganic base KOH to alkaline pH14 (using aqueous b% =15% inorganic base, consumption C =16.3 g aqueous inorganic base at endpoint)
3) Heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.0 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (3.0-0.697 × 16.3 × 15%)/15 × 100% =8.6%, similarly e2% =4.5%, e3% =2.9%, e4% =2.1%, e5% =1.7%, e6% =1.5%;
6) calculating the percentage content of total organic matter (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and 8.6%, 4.5%, 2.9%, 2.1%, 1.7% and 1.5% as a Y axis, establishing a regression equation of Y = -1.2771X + 8.0200 (R = 0.878), enabling X =1, and calculating the Y value as the e% value and e% = 6.74%. The bamboo vinegar liquid has the water percentage content (g%) = 1-the total organic matter percentage content (e%) =1-6.74% = 93.26%.
Example 6
1) Sampling 6 g of bamboo vinegar (accounting for a =6 g), diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of bamboo vinegar +0 g of distilled water; diluting to 2 times refers to 1/2 × a g of bamboo vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of bamboo vinegar and 2/3 × a g of distilled water, and so on) taking 6 g of each of 6 samples for later use;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7 (using aqueous b% =5% inorganic base, titrated to endpoint consuming C =6.0 grams of aqueous inorganic base)
3) Heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.72 g);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (0.72-0.575 × 6.0 × 5%)/6 × 100% =9.1%, similarly e2% =4.6%, e3% =3.2%, e4% =2.4%, e5% =1.8%, e6% =1.6%;
6) calculating the percentage content of total organic matters (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 9.1%, 4.6%, 3.2%, 2.4%, 1.8% and 1.6% as a Y axis, establishing a regression equation Y = -1.3343X + 8.4533 (R = 0.8838), enabling X =1, and calculating the Y value as the e% value e% = 7.12%. The bamboo vinegar liquid has the water content (g%) = 1-total organic matter content (e%) =1-7.12% = 92.88%.
Example 7
1) Sampling 60 g of bamboo vinegar (counting a =60 g) for standby;
2) titration of 50% aqueous inorganic base KOH to alkaline PH12, (using b% =50% aqueous inorganic base, consumption of C =15.2 grams aqueous inorganic base at titration endpoint);
3) heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =9.9 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; e% = (d-0.697 bc%)/a × 100% = (9.9-0.697 × 15.2 × 50%)/60 × 100% =7.9%, = bamboo vinegar moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-7.9% = 92.1%.
Example 8
1) Sampling 30 g of bamboo vinegar (counting a =30 g) for standby;
2) titrating a 50% aqueous inorganic base NaOH to an alkaline PH11, (using b% =50% aqueous inorganic base, titrating to endpoint consumes C =7.5 grams aqueous inorganic base);
3) heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.7 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage contents of the total organic matters of the obtained sample are respectively as follows in sequence: e% = (d-0.575 bc%)/a × 100% = (4.7-0.575 × 7.5 × 50%)/30 × 100% =8.5%, = bamboo vinegar water percentage content (g%) = 1-total organic matter percentage content (e%) =1-8.5% = 91.5%.
Example 9
1) Sampling 30 g of bamboo vinegar (counting a =30 g) for standby;
2) titration of 25% aqueous inorganic base KOH to alkaline PH10, (using b% =25% aqueous inorganic base, consumption of C =14.8 grams aqueous inorganic base at endpoint);
3) heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =5.2 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage contents of the total organic matters of the obtained sample are respectively as follows: e% = (d-0.697 bc%)/a × 100% = (5.2-0.697 × 14.8 × 25%)/30 × 100% =8.7%, the bamboo vinegar water percentage content (g%) = 1-total organic matter percentage content (e%) =1-8.7% = 91.3%.
Example 10
1) Sampling 30 g of bamboo vinegar (counting a =30 g) for standby;
2) aqueous NaOH 20% inorganic base was titrated to alkaline PH9, (b% =20% aqueous inorganic base was used, C =11.2 g aqueous inorganic base was consumed at the end point of the titration);
3) heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.0 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.575 bc%)/a × 100% = (4.0-0.575 × 11.2 × 20%)/30 × 100% = 9.0%. The bamboo vinegar liquid has the water content (g%) = 1-total organic matter content (e%) =1-9.0% = 91.0%.
Example 11
1) Sampling 15 g of bamboo vinegar (counting a =15 g) for standby;
2) titration of aqueous 15% inorganic base KOH to alkaline PH14, (using aqueous b% =15% inorganic base, consumption of aqueous C =16.3 grams of inorganic base at endpoint);
3) heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.0 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (3.0-0.697 × 16.3 × 15%)/15 × 100% =8.6%, = bamboo vinegar moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-8.6% = 92.2%.
Example 12
1) Sampling 6 g of bamboo vinegar (counting a =6 g) for standby;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7, (using b% =5% aqueous inorganic base, titrated to endpoint consuming C =6.0 grams aqueous inorganic base);
3) heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.72 g);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage contents of the total organic matters of the obtained sample are respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (0.72-0.575 × 6.0 × 5%)/6 × 100% =9.1%, = bamboo vinegar water content (% g%) = 1-total organic matter content (% e%) =1-9.1% = 90.9%.
Example 13
1) 60 g of pyroligneous liquor was sampled (a =60 g), and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 60 g of each of 6 samples for later use;
2) titration of 50% aqueous inorganic base KOH to alkaline pH12 (using b% =50% aqueous inorganic base, titration to endpoint consumes C =18.0 grams aqueous inorganic base)
3) Heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =11.9 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (11.9-0.697 × 18.0 × 50%)/60 × 100% =9.4%, similarly e2% =4.8%, e3% =3.2%, e4% =2.4%, e5% =1.9%, e6% =1.6%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 9.4%, 4.8%, 3.2%, 2.4%, 1.9% and 1.6% as a Y axis, establishing a regression equation of Y = -1.3857X + 8.7333 (R2 = 0.7799), enabling X =1, and calculating the Y value as the e% value e% = 7.35%. The percentage content of water (g%) = 1-total organic matter percentage content (e%) =1-7.35% =92.65% of the pyroligneous liquor.
Example 14
1) 30 g of pyroligneous liquor was sampled (a =30 g in terms of a), and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 30 g of each of 6 samples for later use;
2) titration of aqueous NaOH, 50% inorganic base, to alkaline pH11 (b% =50% aqueous inorganic base used, consumption C =5.7 g aqueous inorganic base at endpoint)
3) Heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.7 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (5.7-0.575 × 9.1 × 50%)/30 × 100% =10.3%, similarly e2% =5.2%, e3% =3.4%, e4% =2.6%, e5% =2.0%, e6% =1.8%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and taking 10.3%, 5.2%, 3.4%, 2.6%, 2.0% and 1.8% as a Y axis, establishing a regression equation of Y = Y = -1.5114X + 9.5067 (R2 = 1), and making X =1, and calculating the Y value as the e% value e% = 8.00%; the percentage content of water (g%) = 1-total organic matter percentage content (e%) =1-8.00% = 92.00%.
Example 15
1) 30 g of pyroligneous liquor was sampled (a =30 g in terms of a), and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 30 g of each of 6 samples for later use;
2) titration of aqueous 25% inorganic base KOH to alkaline pH10 (using aqueous b% =25% inorganic base, consumption C =17.8 g aqueous inorganic base at endpoint)
3) Heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =6.3 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (6.3-0.697 × 17.8 × 25%)/30 × 100% =10.7%, similarly e2% =5.4%, e3% =3.6%, e4% =2.7%, e5% =2.2%, e6% =1.8%;
6) the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample were calculated by taking 1, 2, 3, 4, 5, 6 as X-axis, and 10.7%, 5.4%, 3.6%, 2.7%, 2.2%, 1.8% as Y-axis, establishing regression equation Y = -1.5429X + 9.7667 (R2 = 1), making X =1, and calculating Y value as e% value e% = 8.22%. The percentage content of water (g%) = 1-total organic matter percentage content (e%) =1-8.22% =91.78% in the pyroligneous liquor.
Example 16
1) 30 g of pyroligneous liquor was sampled (a =30 g in terms of a), and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 30 g of each of 6 samples for later use;
2) aqueous NaOH 20% inorganic base titrated to alkaline PH9 (using aqueous b% =20% inorganic base, consumption C =13.4 g aqueous inorganic base at endpoint)
3) Heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.8 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (4.8-0.575 × 13.4 × 20%)/30 × 100% =10.9%, similarly e2% =5.5%, e3% =3.6%, e4% =2.7%, e5% =2.1%, e6% =1.8%;
6) calculating the percentage content of total organic matters (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 10.9%, 5.5%, 3.6%, 2.7%, 2.1% and 1.8% as a Y axis, establishing a regression equation of Y = -1.6171X + 10.093 (R2 = 0.999), making X =1, and calculating the Y value as the e% value e% = 8.48%. The percentage content of water (g%) = 1-total organic matter percentage content (e%) =1-8.48% =91.52% in the pyroligneous liquor.
Example 17
1) 15 g of pyroligneous liquor (a =15 g) was sampled, and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 15 g of each of 6 samples for later use;
2) titration of aqueous 15% inorganic base KOH to alkaline pH14 (using aqueous b% =15% inorganic base, consumption C =19.6 g aqueous inorganic base at endpoint)
3) Heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.7 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (3.7-0.697 × 19.6 × 15%)/15 × 100% =11.0%, similarly e2% =5.6%, e3% =3.8%, e4% =2.9%, e5% =2.3%, e6% =1.8%;
6) calculating the percentage content of total organic matter (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and taking 11.0%, 5.6%, 3.8%, 2.9%, 2.3% and 1.8% as a Y axis, establishing a regression equation of Y = -1.6229X + 10.24 (R2 = 0.998), and letting X =1, calculating the Y value as the e% value e% = 8.62%. The percentage content of water (g%) = 1-total organic matter percentage content (e%) =1-8.62% =91.38% of pyroligneous liquor.
Example 18
1) 6 g of pyroligneous liquor (a =6 g) was sampled, and 6 samples were diluted 1-fold, 2-fold, 3-fold, 4-fold, 5-fold, and 6-fold with distilled water (the dilution to 1-fold means a g of pyroligneous liquor +0 g of distilled water; diluting to 2 times means 1/2 × a g of pyroligneous liquor and 1/2 × a g of distilled water, diluting to 3 times means 1/3 × a g of pyroligneous liquor and 2/3 × a g of distilled water, and so on) taking 6 g of each of 6 samples for later use;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7 (using aqueous b% =5% inorganic base, titrated to endpoint consuming C =7.3 grams of aqueous inorganic base)
3) Heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.88 g);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (0.88-0.575 × 7.3 × 5%)/6 × 100% =11.2%, similarly e2% =5.6%, e3% =3.7%, e4% =2.9%, e5% =2.2%, e6% =1.9%;
6) calculating the percentage content of total organic matter (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and taking 11.2%, 5.6%, 3.7%, 2.9%, 2.2% and 1.9% as a Y axis, establishing a regression equation of Y = -1.6429X + 10.333 (R2 = 0.9838), enabling X =1, and calculating the Y value as the e% value, e% = 8.69%. The percentage content of water of the pyroligneous liquor (g%) = 1-total organic matter percentage content (e%) =1-8.69% = 91.31%.
Example 19
1) 60 g of pyroligneous liquor is sampled (counting a =60 g) for standby;
2) titration of 50% aqueous inorganic base KOH to alkaline pH12 (using b% =50% aqueous inorganic base, titration to endpoint consumes C =18.0 grams aqueous inorganic base)
3) Heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =11.9 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (11.9-0.697 × 18.0 × 50%)/60 × 100% =9.4%,% moisture content of pyroligneous liquor (g%) = 1-total organic matter content (e%) =1-9.4% = 90.6%.
Example 20
1) Sampling 30 g of pyroligneous liquor (counting for a =30 g) for standby;
2) titration of aqueous NaOH, 50% inorganic base, to alkaline pH11 (b% =50% aqueous inorganic base used, consumption C =5.7 g aqueous inorganic base at endpoint)
3) Heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.7 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.575 bc%)/a × 100% = (5.7-0.575 × 9.1 × 50%)/30 × 100% =10.3%, = pyroligneous liquor moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-10.3% = 89.7%.
Example 21
1) Sampling 30 g of pyroligneous liquor (counting for a =30 g) for standby;
2) titration of aqueous 25% inorganic base KOH to alkaline pH10 (using aqueous b% =25% inorganic base, consumption C =17.8 g aqueous inorganic base at endpoint)
3) Heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =6.3 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (6.3-0.697 × 17.8 × 25%)/30 × 100% =10.7%,% moisture content of pyroligneous liquor (g%) = 1-total organic matter content (e%) =1-10.7% = 89.3%.
Example 22
1) Sampling 30 g of pyroligneous liquor (counting for a =30 g) for standby;
2) aqueous NaOH 20% inorganic base titrated to alkaline PH9 (using aqueous b% =20% inorganic base, consumption C =13.4 g aqueous inorganic base at endpoint)
3) Heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.8 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the sample is obtained, i.e% = (d-0.575 bc%)/a × 100% = (4.8-0.575 × 13.4 × 20%)/30 × 100% =10.9%, and the pyroligneous liquor water percentage content (g%) = 1-total organic matter percentage content (e%) =1-10.9% = 89.1%.
Example 23
1) Sampling 15 g of pyroligneous liquor (accounting for a =15 g) for standby;
2) titration of aqueous 15% inorganic base KOH to alkaline pH14 (using aqueous b% =15% inorganic base, consumption C =19.6 g aqueous inorganic base at endpoint)
3) Heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.7 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is as follows: e% = (d-0.697 bc%)/a × 100% = (3.7-0.697 × 19.6 × 15%)/15 × 100% =11.0%, = pyroligneous liquor moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-11.0% = 89.0%.
Example 24
1) Sampling 6 g of pyroligneous liquor (counting as a =6 g) for standby;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7 (using aqueous b% =5% inorganic base, titrated to endpoint consuming C =7.3 grams of aqueous inorganic base)
3) Heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.88 g);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; obtaining the total organic matter percentage content of 6 samples: e% = (d-0.575 bc%)/a × 100% = (0.88-0.575 × 7.3 × 5%)/6 × 100% =11.2%, = pyroligneous liquor moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-11.2% = 88.8%.
Example 25
1) Sampling 60 g of the grass vinegar (the amount is a =60 g), and diluting the grass vinegar with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of the grass vinegar and 1/2 × a g of the distilled water, diluting to 3 times refers to 1/3 × a g of the grass vinegar and 2/3 × a g of the distilled water, and so on) respectively taking 60 g of 6 samples for later use;
2) titration of 50% aqueous inorganic base KOH to alkaline pH12 (using b% =50% aqueous inorganic base, titration to endpoint consumes C =16.2 grams aqueous inorganic base)
3) Heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =10.9 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (10.9-0.697 × 16.2 × 50%)/60 × 100% =8.8%, similarly e2% =4.3%, e3% =3.0%, e4% =2.3%, e5% =1.8%, e6% =1.6%;
6) calculating the percentage content (e%) and the percentage content (g%) of the total organic matters of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 8.8%, 4.3%, 3.0%, 2.3%, 1.5% and 1.6% as a Y axis, establishing a regression equation Y = -1.2629X + 8.0533
(R2 = 0.9799), let X =1, and the calculated Y value is the e% value e% = 6.79%. The percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-6.79% = 93.21%.
Example 26
1) Sampling 30 g of the grass vinegar (accounting for a =30 g), and diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of grass vinegar liquid and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of grass vinegar liquid and 2/3 × a g of distilled water, and so on) respectively taking 30 g of 6 samples for later use;
2) titration of aqueous NaOH, 50% inorganic base, to alkaline pH11 (b% =50% aqueous inorganic base used, consumption C =5.1 g aqueous inorganic base at endpoint)
3) Heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.2 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (4.2-0.575 × 5.1 × 50%)/30 × 100% =9.1%, similarly e2% =4.6%, e3% =3.1%, e4% =2.4%, e5% =1.9%, e6% =1.5%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as X axes and taking 9.1%, 4.6%, 3.1%, 2.4%, 1.9% and 1.5% as Y axes, establishing a regression equation Y = -1.3371X + 8.4467 (R2 = 1), enabling X =1, and calculating the Y value as the e% value e% = 7.11%; the percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-7.11% = 92.89%.
Example 27
1) Sampling 30 g of the grass vinegar (accounting for a =30 g), and diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of grass vinegar liquid and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of grass vinegar liquid and 2/3 × a g of distilled water, and so on) respectively taking 30 g of 6 samples for later use;
2) titration of aqueous 25% inorganic base KOH to alkaline pH10 (using aqueous b% =25% inorganic base, consumption C =16.0 g aqueous inorganic base at endpoint)
3) Heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =5.8 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (5.8-0.697 × 16.0 × 25%)/30 × 100% =10.4%, similarly e2% =5.6%, e3% =3.5%, e4% =2.7%, e5% =2.1%, e6% =1.7%;
6) calculating the total organic matter percentage content (e%) and the moisture percentage content (g%) of the sample by taking 1, 2, 3, 4, 5 and 6 as X axes and taking 10.4%, 5.6%, 3.5%, 2.7%, 2.1% and 1.7% as Y axes, establishing a regression equation of Y = -1.5657X + 9.8133 (R2 = 1), enabling X =1, and calculating the Y value as the e% value of e% = 8.25%. The percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-8.25% = 91.75%.
Example 28
1) Sampling 30 g of the grass vinegar (accounting for a =30 g), and diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of grass vinegar liquid and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of grass vinegar liquid and 2/3 × a g of distilled water, and so on) respectively taking 30 g of 6 samples for later use;
2) aqueous NaOH 20% inorganic base titrated to alkaline PH9 (using aqueous b% =20% inorganic base, consumption C =12.1 g aqueous inorganic base at endpoint)
3) Heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.4 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (4.4-0.575 × 12.1 × 20%)/30 × 100% =9.6%, similarly e2% =4.9%, e3% =3.3%, e4% =2.6%, e5% =2.0%, e6% =1.6%;
6) calculating the percentage content of total organic matters (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis, 9.6%, 4.9%, 3.3%, 2.6%, 2.0% and 1.6% as a Y axis, establishing a regression equation of Y = Y = -1.4114X + 8.94 (R2 = 0.989), making X =1, and calculating the Y value as the e% value e% = 7.53%. The percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-7.53% = 92.47%.
Example 29
1) Sampling 15 g of the grass vinegar (accounting for a =15 g), and diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of the grass vinegar and 1/2 × a g of the distilled water, diluting to 3 times refers to 1/3 × a g of the grass vinegar and 2/3 × a g of the distilled water, and so on) taking 15 g of each of 6 samples for later use;
2) titration of aqueous 15% inorganic base KOH to alkaline pH14 (using aqueous b% =15% inorganic base, consumption C =17.6 g aqueous inorganic base at endpoint)
3) Heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.4 grams);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.697 bc%)/a × 100% = (3.4-0.697 × 17.6 × 15%)/15 × 100% =10.4%, similarly e2% =5.8%, e3% =3.5%, e4% =2.6%, e5% =2.1%, e6% =1.7%;
6) calculating the percentage content of total organic matter (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and taking 10.4%, 5.8%, 3.5%, 2.6%, 2.1% and 1.7% as a Y axis, establishing a regression equation Y = -1.5857X + 9.9 (R2 = 0.988), making X =1, and calculating the Y value as the e% value e% = 8.31%. The percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-8.31% = 91.69%.
Example 30
1) Sampling 6 g of the grass vinegar (accounting for a =6 g), and diluting with distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times to obtain 6 samples (the dilution to 1 time means a g of the grass vinegar and 0 g of the distilled water; diluting to 2 times refers to 1/2 × a g of grass vinegar and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of grass vinegar and 2/3 × a g of distilled water, and so on) taking 6 g of each of 6 samples for later use;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7 (using aqueous b% =5% inorganic base, titrated to endpoint consuming C =6.6 grams of aqueous inorganic base)
3) Heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.80 g);
5) the total organic matter percentage content (e%) of 6 samples was calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1% = (d-0.575 bc%)/a × 100% = (0.80-0.575 × 6.6 × 5%)/6 × 100% =10.2%, similarly e2% =5.7%, e3% =3.4%, e4% =2.7%, e5% =2.0%, e6% =1.7%;
6) calculating the percentage content of total organic matter (e%) and the percentage content of moisture (g%) of the sample, taking 1, 2, 3, 4, 5 and 6 as an X axis and taking 10.2%, 5.7%, 3.4%, 2.7%, 2.0% and 1.7% as a Y axis, establishing a regression equation of Y = -1.5514X + 9.7133 (R2 = 0.9838), enabling X =1, and calculating the Y value as the e% value e% = 8.16%. The percentage content of water in the herbal vinegar solution (g%) = 1-total organic matter percentage content (e%) =1-8.16% = 91.84%.
Example 31
1) Sampling 60 g of the herbal vinegar (counting a =60 g) for standby;
2) titration of 50% aqueous inorganic base KOH to alkaline pH12 (using b% =50% aqueous inorganic base, titration to endpoint consumes C =16.2 grams aqueous inorganic base)
3) Heating to 150 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =10.9 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (10.9-0.697 × 16.2 × 50%)/60 × 100% =8.8%, = water content of the herbal and vinegar solution (g%) = 1-total organic matter content (e%) =1-6.79% = 93.21%.
Example 32
1) Sampling 30 g of the herbal vinegar (counting a =30 g) for standby;
2) titration of aqueous NaOH, 50% inorganic base, to alkaline pH11 (b% =50% aqueous inorganic base used, consumption C =5.1 g aqueous inorganic base at endpoint)
3) Heating to 140 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.2 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.575 bc%)/a × 100% = (4.2-0.575 × 5.1 × 50%)/30 × 100% =9.1%, = the percentage moisture content of the herbal vinegar (g%) = 1-the total organic matter content (e%) =1-9.1% = 91.9%.
Example 33
1) Sampling 30 g of the herbal vinegar (counting a =30 g) for standby;
2) titration of aqueous 25% inorganic base KOH to alkaline pH10 (using aqueous b% =25% inorganic base, consumption C =16.0 g aqueous inorganic base at endpoint)
3) Heating to 120 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =5.8 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (5.8-0.697 × 16.0 × 25%)/30 × 100% =10.4%, = the percentage of moisture content of the herbal and vinegar solution (g%) = 1-total organic matter content (e%) =1-10.4% = 89.6%.
Example 34
1) Sampling 30 g of the herbal vinegar (counting a =30 g) for standby;
2) aqueous NaOH 20% inorganic base titrated to alkaline PH9 (using aqueous b% =20% inorganic base, consumption C =12.1 g aqueous inorganic base at endpoint)
3) Heating to 110 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =4.4 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.575 bc%)/a × 100% = (4.4-0.575 × 12.1 × 20%)/30 × 100% =9.6%, = the percentage moisture content of the herbal vinegar (g%) = 1-the total organic matter content (e%) =1-9.6% = 90.4%.
Example 35
1) Sampling 15 g of the herbal vinegar (counting a =15 g) for standby;
2) titration of aqueous 15% inorganic base KOH to alkaline pH14 (using aqueous b% =15% inorganic base, consumption C =17.6 g aqueous inorganic base at endpoint)
3) Heating to 100 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =3.4 grams);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.697 bc%)/a × 100% = (3.4-0.697 × 17.6 × 15%)/15 × 100% =10.4%, = straw vinegar moisture percentage content (g%) = 1-total organic matter percentage content (e%) =1-10.4% = 89.6%.
Example 36
1) Sampling 6 g of the herbal vinegar (counting a =6 g) for standby;
2) aqueous 5% inorganic base NaOH titrated to alkaline PH7 (using aqueous b% =5% inorganic base, titrated to endpoint consuming C =6.6 grams of aqueous inorganic base)
3) Heating to 130 deg.C, and drying to completely volatilize the liquid;
4) the total mass of the remaining material was weighed (d =0.80 g);
5) the percentage content of total organic matters (e%) and the percentage content of water (g%) of the sample are calculated by the following method: using NaOH: e% = (d-0.575 bc%)/a × 100%; the percentage content of the total organic matters of the obtained sample is as follows: e% = (d-0.575 bc%)/a × 100% = (0.80-0.575 × 6.6 × 5%)/6 × 100% =10.2%, and the water content of the herbal and vinegar solution (g%) = 1-total organic matter content (e%) =1-10.2% = 89.8%.
The following test shows the effect of the above-described method of the present invention.
In table 1, the negative number of errors is smaller than the actual value, the positive number is larger than the actual value, and the results are sorted according to the absolute value of the errors from small to large in the effect sorting, and the data show that the errors of the percentage contents of the machine mass of the bamboo vinegar in examples 1-12 are all smaller than 2%, the accuracy sorting of the regression method used in examples 1-6 is 1-7 leading lines, compared with the actual measurement method in examples 7-12, the regression method is more accurate than the actual measurement method, wherein the error of example 3-5 before sorting is 3, the description accuracy is higher, the best error of example 4 is only 0.2%, compared with the error of 1.8% in example 10 of the actual measurement method, the total error reaches 2.0%, and the regression method is better than the actual measurement method. The data also show that the percentage content of organic matters in the bamboo vinegar liquid can be measured in examples 1-12, the percentage content of moisture can also be measured, errors are less than 2% when the moisture content of the sample reaches more than 90%, and the accuracy is very high.
In table 2, the negative number of errors is smaller than the actual value, the positive number is larger than the actual value, and the results are sorted according to the absolute value of the errors from small to large in the effect sorting, and the data show that the errors of the percentage contents of the pyroligneous liquor in examples 13-24 are all smaller than 2.5%, the accuracy sorting of the regression method used in examples 13-18 is 13-19 front rows, compared with the actual measurement method in examples 19-24, the regression method is more accurate than the actual measurement method, wherein the error of the best example 18 is only 0.1% in 3 before the sorting of examples 16-18, the accuracy is higher, the total error is 2.5% compared with the error of the corresponding actual measurement method, namely 2.4% in example 24, and the regression method is better than the actual measurement method. The data also show that the percentage content of organic matters in the pyroligneous liquor can be measured in examples 13-24, the percentage content of water can also be measured, errors are less than 2.5% when the moisture content of the sample reaches more than 88%, and the accuracy is very high.
In table 3, the negative number of errors is smaller than the actual value, the positive number is larger than the actual value, and the results are sorted according to the absolute value of the errors from small to large in the effect sorting, and the data shows that the errors of the mass percentage of the herbal vinegar liquid in examples 25-36 are all less than 2.5%, the accuracy sorting of the regression method used in examples 25-30 is 25-28 front rows, and compared with the actual measurement method in examples 31-36, the regression method is more accurate than the actual measurement method, wherein the error of example 30\26\29 is higher in description accuracy, the best error of example 30 is only 0.16%, and compared with the error of 2.4% in example 35 of the actual measurement method, the total error reaches 2.56%, and the regression method is better than the actual measurement method. The data also show that the percentage content of organic matters in the grass vinegar liquid can be measured in examples 25 to 36, the percentage content of water can also be measured, errors are less than 2.5% when the moisture content of a sample reaches more than 89%, and the accuracy is very high.
Claims (2)
1. A method for conveniently measuring the total organic matter and water content of plant vinegar is characterized by comprising the following steps:
1) sampling 6-60 g of the plant vinegar, calculating a g, diluting 6 samples by using distilled water to 1 time, 2 times, 3 times, 4 times, 5 times and 6 times, wherein the dilution to 1 time means a g of plant vinegar and 0 g of distilled water; diluting to 2 times refers to 1/2 × a g of plant vinegar liquid and 1/2 × a g of distilled water, diluting to 3 times refers to 1/3 × a g of plant vinegar liquid and 2/3 × a g of distilled water, and so on, and taking a g of samples for later use;
2) titrating the plant vinegar liquid diluted in the step 1) with 5-50% of inorganic base aqueous solution until the alkaline pH value is 7-14, wherein b% of inorganic base aqueous solution is used, and c g of inorganic base aqueous solution is consumed when the titration is finished; the inorganic alkaline water solution is NaOH water solution or KOH water solution;
3) heating the plant vinegar liquid treated in the step 2) to 100-150 ℃ or drying by hot air to completely volatilize the liquid;
4) weighing the total mass of the rest substances, and calculating d grams;
5) the percentage content e% of total organic matter of 6 samples is calculated, and the calculation method comprises the following steps: using NaOH: e% = (d-0.575 bc%)/a × 100%; using KOH: e% = (d-0.697 bc%)/a × 100%; the total organic matter percentage content of the obtained 6 samples is respectively as follows in sequence: e1%, e2%, e3%, e4%, e5%, e 6%;
6) calculating the percentage content e% of total organic matters and the percentage content g% of water of the plant vinegar liquid, taking 1, 2, 3, 4, 5 and 6 as an X axis, taking e1%, e2%, e3%, e4%, e5% and e6% as a Y axis, establishing a regression equation, enabling X =1, and calculating a Y value to be an e% value; the percentage content of water in the plant vinegar liquid is g percent, g percent is not less than 1 percent, and the total organic matter percentage content of the plant vinegar liquid is e percent;
the plant vinegar is one of bamboo vinegar, wood vinegar or grass vinegar.
2. A method for conveniently measuring the total organic matter and water content of plant vinegar is characterized by comprising the following steps:
1) sampling 6-60 g of plant vinegar liquid, and counting as a g for later use;
2) titrating the plant vinegar liquid obtained in the step 1) with 5-50% of inorganic base aqueous solution until the alkaline pH value is 7-14, wherein the b% of inorganic base aqueous solution is used, and c g of inorganic base aqueous solution is consumed when the titration is finished; the inorganic alkaline water solution is NaOH water solution or KOH water solution;
3) heating the plant vinegar liquid treated in the step 2) to the temperature of 100-110 ℃ or drying by hot air to completely volatilize the liquid;
4) weighing the total mass of the rest substances, and calculating d grams;
5) the percentage content e% of total organic matter and the percentage content g% of water of the plant vinegar liquid are calculated, and the calculation method comprises the following steps: using NaOH: e% = (d-0.575 bc%)/a × 100%; using KOH: e% = (d-0.697 bc%)/a × 100%; the percentage content of the total organic matters of the plant vinegar liquid is e percent; the percentage content of water in the plant vinegar liquid is g percent, g percent is not less than 1 percent, and the total organic matter percentage content of the plant vinegar liquid is e percent;
the plant vinegar is one of bamboo vinegar, wood vinegar or grass vinegar.
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| CN201810081576.3A CN108333019A (en) | 2018-01-25 | 2018-01-25 | A kind of convenient method for measuring wood vinegar total organic matter and moisture |
| CN2018100815763 | 2018-01-25 | ||
| CN201810081580X | 2018-01-25 | ||
| CN2018100815778 | 2018-01-25 | ||
| CN201810081577.8A CN108254282A (en) | 2018-01-25 | 2018-01-25 | A kind of convenient method for measuring careless vinegar liquid total organic matter and moisture |
| CN201810081580.XA CN108287120A (en) | 2018-01-25 | 2018-01-25 | A kind of convenient method for measuring bamboo vinegar total organic matter and moisture |
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| CN106635107A (en) * | 2016-11-21 | 2017-05-10 | 浙江农林大学 | Concentrating and tar removing method of biomass gasification liquid (bamboo vinegar) |
| CN106680419B (en) * | 2016-12-12 | 2020-12-25 | 浙江农林大学 | Method for conveniently determining contents of organic acid and grass tar in grass vinegar |
| CN106769601B (en) * | 2016-12-12 | 2020-10-23 | 浙江农林大学 | Method for conveniently determining content of organic acid and wood tar in wood vinegar |
| CN106596324B (en) * | 2016-12-12 | 2020-10-23 | 浙江农林大学 | Method for conveniently determining content of organic acid and bamboo tar in bamboo vinegar |
| CN108333019A (en) * | 2018-01-25 | 2018-07-27 | 浙江农林大学 | A kind of convenient method for measuring wood vinegar total organic matter and moisture |
| CN108254282A (en) * | 2018-01-25 | 2018-07-06 | 浙江农林大学 | A kind of convenient method for measuring careless vinegar liquid total organic matter and moisture |
| CN108287120A (en) * | 2018-01-25 | 2018-07-17 | 浙江农林大学 | A kind of convenient method for measuring bamboo vinegar total organic matter and moisture |
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