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How could I prepare acidified methanol solution?
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Nadine Abu Omar
How could I prepare acidified methanol solution?
If you want not to have any water, apart from bubbling in HCl vapor as suggested by Ruben Leitao, you can also make "methanolic hydrogen chloride (NOT methanolic hydrochloric acid)" as described in this reference:
"One popular reagent is methanolic hydrogen chloride (NOT methanolic hydrochloric acid), and it is most easily prepared by adding acetyl chloride (5 mL) slowly to cooled dry methanol (50 mL). Methyl acetate is formed as a by-product, but it does not interfere with methylations at this concentration. A solution of 1-2% (v/v) concentrated sulfuric acid in methanol trans-esterifies lipids in the same manner and at much the same rate; it is very easy to prepare whenever it is required simply by carefully pipetting the appropriate amount of acid into cold methanol. Fresh reagent is best, but if need be it can be stored refrigerated for up to 4 weeks." https://www.lipidhome.co.uk/ms/basics/msmeprep/index.htm ArticleAcetyl Chloride—Methanol as a Convenient Reagent for: A) Qua...
If you want not to have any water, apart from bubbling in HCl vapor as suggested by Ruben Leitao, you can also make "methanolic hydrogen chloride (NOT methanolic hydrochloric acid)" as described in this reference:
"One popular reagent is methanolic hydrogen chloride (NOT methanolic hydrochloric acid), and it is most easily prepared by adding acetyl chloride (5 mL) slowly to cooled dry methanol (50 mL). Methyl acetate is formed as a by-product, but it does not interfere with methylations at this concentration. A solution of 1-2% (v/v) concentrated sulfuric acid in methanol trans-esterifies lipids in the same manner and at much the same rate; it is very easy to prepare whenever it is required simply by carefully pipetting the appropriate amount of acid into cold methanol. Fresh reagent is best, but if need be it can be stored refrigerated for up to 4 weeks." https://www.lipidhome.co.uk/ms/basics/msmeprep/index.htm ArticleAcetyl Chloride—Methanol as a Convenient Reagent for: A) Qua...
Hi Song! Recently I have used the acidified methanol as per following; 1. 10% glacial acetic acid in methanol by adding 10 ml G. acetic acid + 90 ml Methanol AR. 2. 0.3% alcoholic HCl by taking 49.3 ml methanol AR + 0.3 ml concentrated HCl. Mostly, in research papers it is not mentioned but once I was in international conference & i have asked the same question to the one of the scientist from Germany, He told me that, IF IT IS NOT MENTIONED HOW MUCH CONCENTRATION TO TAKE, THEN TAKE CONCENTRATED WHICH IS SOLD COMMERCIALLY. This is what my input, might be other researchers have different ideas.
Hi Song! Recently I have used the acidified methanol as per following; 1. 10% glacial acetic acid in methanol by adding 10 ml G. acetic acid + 90 ml Methanol AR. 2. 0.3% alcoholic HCl by taking 49.3 ml methanol AR + 0.3 ml concentrated HCl. Mostly, in research papers it is not mentioned but once I was in international conference & i have asked the same question to the one of the scientist from Germany, He told me that, IF IT IS NOT MENTIONED HOW MUCH CONCENTRATION TO TAKE, THEN TAKE CONCENTRATED WHICH IS SOLD COMMERCIALLY. This is what my input, might be other researchers have different ideas.
As far as I know, HCl is usually not used in the acidification of methanol & it is even avoided. Instead, H2SO4 or H3PO4 are used to get acidified methanol. By stating 80% acidified...etc., the article made another mistake since the figure indicates the percentage of the acid used while the ratio gives it to methanol. The third mistake is that HCl already contains water (it is called hydrochloric acid) & the concentrated is normally 32% or 36% which means a weight content of water of 68% or 64%. To prepare 20% acidified CH3OH solution, use 24 g. sulfuric acid & mix it with 76 g methanol (if conc. H2SO4 is 96%).
As far as I know, HCl is usually not used in the acidification of methanol & it is even avoided. Instead, H2SO4 or H3PO4 are used to get acidified methanol. By stating 80% acidified...etc., the article made another mistake since the figure indicates the percentage of the acid used while the ratio gives it to methanol. The third mistake is that HCl already contains water (it is called hydrochloric acid) & the concentrated is normally 32% or 36% which means a weight content of water of 68% or 64%. To prepare 20% acidified CH3OH solution, use 24 g. sulfuric acid & mix it with 76 g methanol (if conc. H2SO4 is 96%).
Given this specific context, I would say that «80% acidified methanol» (rather ambiguously written) seems to intentionally refer to a prior 80% v/v methanol, aq. sol., then acidified by adding a small quantity of conc. HCl. That being the case, it should have been clearly stated that the percentage concentration should be taken as volumetric (e.g. % v/v should substitute for %).
Not considering this specific context, % concentrations, namely those indicated in commercial labels, should generically be assumed as given in wt%, unless otherwise stated.
For 80% v/v methanol, it should be understood that 80 vol. were added to water, until the solution fills 100 vol.; each vol. is some arbitrary volume (e.g. 1 cm3).
That is incongruent, however, with the formulation volume ratio given for that same solution: «HCl/methanol/water, 1:80:10, v/v» (not unambiguously written) should be read as 'conc. HCl/methanol/water, 1:80:10, v/v/v'. This is compatible with a prior unacidified methanol aq. sol., prepared after a 80:10 (v/v) formulation ratio, which is obviously not an 80% v/v methanol aq. sol.
It should be noticed that volumetric formulation ratios are not to be confused with volumetric concentrations. Moreover, volume is not conserved while preparing the solution.
Given this specific context, I would say that «80% acidified methanol» (rather ambiguously written) seems to intentionally refer to a prior 80% v/v methanol, aq. sol., then acidified by adding a small quantity of conc. HCl. That being the case, it should have been clearly stated that the percentage concentration should be taken as volumetric (e.g. % v/v should substitute for %).
Not considering this specific context, % concentrations, namely those indicated in commercial labels, should generically be assumed as given in wt%, unless otherwise stated.
For 80% v/v methanol, it should be understood that 80 vol. were added to water, until the solution fills 100 vol.; each vol. is some arbitrary volume (e.g. 1 cm3).
That is incongruent, however, with the formulation volume ratio given for that same solution: «HCl/methanol/water, 1:80:10, v/v» (not unambiguously written) should be read as 'conc. HCl/methanol/water, 1:80:10, v/v/v'. This is compatible with a prior unacidified methanol aq. sol., prepared after a 80:10 (v/v) formulation ratio, which is obviously not an 80% v/v methanol aq. sol.
It should be noticed that volumetric formulation ratios are not to be confused with volumetric concentrations. Moreover, volume is not conserved while preparing the solution.
Hi Song-Lei, Unless explicitly mentioned, I would take it to be concentrated HCl, which is about 12N. Based on what you mention 1:80:10 it would bring it to about a 132mM HCl in the final solution. the 1:80:10 does not match with the 80% Methanol as the 1:80:10 would mean it is about 88% Methanol. Anyway I have never used a 80% acidified Methanol. But what I usually use acidified methanol (acidified with HCl) is for dissolving coelenterazine. What I use is 200 uL of 3N HCl in 9.8 ml of Methanol (every 10 ml of acidified Methanol contains 0.2 ml of 3N HCl).
Hi Song-Lei, Unless explicitly mentioned, I would take it to be concentrated HCl, which is about 12N. Based on what you mention 1:80:10 it would bring it to about a 132mM HCl in the final solution. the 1:80:10 does not match with the 80% Methanol as the 1:80:10 would mean it is about 88% Methanol. Anyway I have never used a 80% acidified Methanol. But what I usually use acidified methanol (acidified with HCl) is for dissolving coelenterazine. What I use is 200 uL of 3N HCl in 9.8 ml of Methanol (every 10 ml of acidified Methanol contains 0.2 ml of 3N HCl).
Wasim H Chowdhury Dr.Wasim H Chowdhury I am a student in analytical chemistry and I want to know about preparing methanolic acid I see your previously answer, and I hope to help me, Are there any references about this preparation Thank you very much.
Wasim H Chowdhury Dr.Wasim H Chowdhury I am a student in analytical chemistry and I want to know about preparing methanolic acid I see your previously answer, and I hope to help me, Are there any references about this preparation Thank you very much.
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
The answer is unclear because the information you have provided is unclear. 1:80:10::HCl:MeOH:water v/v is clear although not a standard description. It means simply to combine 1mL HCl (which if not specified means conc. HCl) with 80mL of methanol and 10 mL of water. This however is not an 80% solution. What is probably meant is 1:80:20 v/v where the total exceeds 100% but that is a common way to describe a system when one of the components is much less than the others, i.e. 1:79.5:19.5 or some variation on this theme is very hard to process in the lab. The respondent talking about H2SO4 is missing the point - you don't need dry acidified methanol when 20% of the solution is water.
The answer is unclear because the information you have provided is unclear. 1:80:10::HCl:MeOH:water v/v is clear although not a standard description. It means simply to combine 1mL HCl (which if not specified means conc. HCl) with 80mL of methanol and 10 mL of water. This however is not an 80% solution. What is probably meant is 1:80:20 v/v where the total exceeds 100% but that is a common way to describe a system when one of the components is much less than the others, i.e. 1:79.5:19.5 or some variation on this theme is very hard to process in the lab. The respondent talking about H2SO4 is missing the point - you don't need dry acidified methanol when 20% of the solution is water.
If you want not to have any water, apart from bubbling in HCl vapor as suggested by Ruben Leitao, you can also make "methanolic hydrogen chloride (NOT methanolic hydrochloric acid)" as described in this reference:
"One popular reagent is methanolic hydrogen chloride (NOT methanolic hydrochloric acid), and it is most easily prepared by adding acetyl chloride (5 mL) slowly to cooled dry methanol (50 mL). Methyl acetate is formed as a by-product, but it does not interfere with methylations at this concentration. A solution of 1-2% (v/v) concentrated sulfuric acid in methanol trans-esterifies lipids in the same manner and at much the same rate; it is very easy to prepare whenever it is required simply by carefully pipetting the appropriate amount of acid into cold methanol. Fresh reagent is best, but if need be it can be stored refrigerated for up to 4 weeks."
https://www.lipidhome.co.uk/ms/basics/msmeprep/index.htm
Article Acetyl Chloride—Methanol as a Convenient Reagent for: A) Qua...
If you want not to have any water, apart from bubbling in HCl vapor as suggested by Ruben Leitao, you can also make "methanolic hydrogen chloride (NOT methanolic hydrochloric acid)" as described in this reference:
"One popular reagent is methanolic hydrogen chloride (NOT methanolic hydrochloric acid), and it is most easily prepared by adding acetyl chloride (5 mL) slowly to cooled dry methanol (50 mL). Methyl acetate is formed as a by-product, but it does not interfere with methylations at this concentration. A solution of 1-2% (v/v) concentrated sulfuric acid in methanol trans-esterifies lipids in the same manner and at much the same rate; it is very easy to prepare whenever it is required simply by carefully pipetting the appropriate amount of acid into cold methanol. Fresh reagent is best, but if need be it can be stored refrigerated for up to 4 weeks."
https://www.lipidhome.co.uk/ms/basics/msmeprep/index.htm
Article Acetyl Chloride—Methanol as a Convenient Reagent for: A) Qua...
More
VOTE
Hi Song! Recently I have used the acidified methanol as per following;
1. 10% glacial acetic acid in methanol by adding 10 ml G. acetic acid + 90 ml Methanol AR.
2. 0.3% alcoholic HCl by taking 49.3 ml methanol AR + 0.3 ml concentrated HCl.
Mostly, in research papers it is not mentioned but once I was in international conference & i have asked the same question to the one of the scientist from Germany, He told me that, IF IT IS NOT MENTIONED HOW MUCH CONCENTRATION TO TAKE, THEN TAKE CONCENTRATED WHICH IS SOLD COMMERCIALLY.
This is what my input, might be other researchers have different ideas.
Hi Song! Recently I have used the acidified methanol as per following;
1. 10% glacial acetic acid in methanol by adding 10 ml G. acetic acid + 90 ml Methanol AR.
2. 0.3% alcoholic HCl by taking 49.3 ml methanol AR + 0.3 ml concentrated HCl.
Mostly, in research papers it is not mentioned but once I was in international conference & i have asked the same question to the one of the scientist from Germany, He told me that, IF IT IS NOT MENTIONED HOW MUCH CONCENTRATION TO TAKE, THEN TAKE CONCENTRATED WHICH IS SOLD COMMERCIALLY.
This is what my input, might be other researchers have different ideas.
More
VOTE
As far as I know, HCl is usually not used in the acidification of methanol & it is even avoided. Instead, H2SO4 or H3PO4 are used to get acidified methanol. By stating 80% acidified...etc., the article made another mistake since the figure indicates the percentage of the acid used while the ratio gives it to methanol. The third mistake is that HCl already contains water (it is called hydrochloric acid) & the concentrated is normally 32% or 36% which means a weight content of water of 68% or 64%. To prepare 20% acidified CH3OH solution, use 24 g. sulfuric acid & mix it with 76 g methanol (if conc. H2SO4 is 96%).
As far as I know, HCl is usually not used in the acidification of methanol & it is even avoided. Instead, H2SO4 or H3PO4 are used to get acidified methanol. By stating 80% acidified...etc., the article made another mistake since the figure indicates the percentage of the acid used while the ratio gives it to methanol. The third mistake is that HCl already contains water (it is called hydrochloric acid) & the concentrated is normally 32% or 36% which means a weight content of water of 68% or 64%. To prepare 20% acidified CH3OH solution, use 24 g. sulfuric acid & mix it with 76 g methanol (if conc. H2SO4 is 96%).
More
VOTE
Given this specific context, I would say that «80% acidified methanol» (rather ambiguously written) seems to intentionally refer to a prior 80% v/v methanol, aq. sol., then acidified by adding a small quantity of conc. HCl. That being the case, it should have been clearly stated that the percentage concentration should be taken as volumetric (e.g. % v/v should substitute for %).
Not considering this specific context, % concentrations, namely those indicated in commercial labels, should generically be assumed as given in wt%, unless otherwise stated.
For 80% v/v methanol, it should be understood that 80 vol. were added to water, until the solution fills 100 vol.; each vol. is some arbitrary volume (e.g. 1 cm3).
That is incongruent, however, with the formulation volume ratio given for that same solution: «HCl/methanol/water, 1:80:10, v/v» (not unambiguously written) should be read as 'conc. HCl/methanol/water, 1:80:10, v/v/v'. This is compatible with a prior unacidified methanol aq. sol., prepared after a 80:10 (v/v) formulation ratio, which is obviously not an 80% v/v methanol aq. sol.
It should be noticed that volumetric formulation ratios are not to be confused with volumetric concentrations. Moreover, volume is not conserved while preparing the solution.
Given this specific context, I would say that «80% acidified methanol» (rather ambiguously written) seems to intentionally refer to a prior 80% v/v methanol, aq. sol., then acidified by adding a small quantity of conc. HCl. That being the case, it should have been clearly stated that the percentage concentration should be taken as volumetric (e.g. % v/v should substitute for %).
Not considering this specific context, % concentrations, namely those indicated in commercial labels, should generically be assumed as given in wt%, unless otherwise stated.
For 80% v/v methanol, it should be understood that 80 vol. were added to water, until the solution fills 100 vol.; each vol. is some arbitrary volume (e.g. 1 cm3).
That is incongruent, however, with the formulation volume ratio given for that same solution: «HCl/methanol/water, 1:80:10, v/v» (not unambiguously written) should be read as 'conc. HCl/methanol/water, 1:80:10, v/v/v'. This is compatible with a prior unacidified methanol aq. sol., prepared after a 80:10 (v/v) formulation ratio, which is obviously not an 80% v/v methanol aq. sol.
It should be noticed that volumetric formulation ratios are not to be confused with volumetric concentrations. Moreover, volume is not conserved while preparing the solution.
More
VOTE
Hi Song-Lei,
Unless explicitly mentioned, I would take it to be concentrated HCl, which is about 12N. Based on what you mention 1:80:10 it would bring it to about a 132mM HCl in the final solution. the 1:80:10 does not match with the 80% Methanol as the 1:80:10 would mean it is about 88% Methanol.
Anyway I have never used a 80% acidified Methanol. But what I usually use acidified methanol (acidified with HCl) is for dissolving coelenterazine. What I use is 200 uL of 3N HCl in 9.8 ml of Methanol (every 10 ml of acidified Methanol contains 0.2 ml of 3N HCl).
Hi Song-Lei,
Unless explicitly mentioned, I would take it to be concentrated HCl, which is about 12N. Based on what you mention 1:80:10 it would bring it to about a 132mM HCl in the final solution. the 1:80:10 does not match with the 80% Methanol as the 1:80:10 would mean it is about 88% Methanol.
Anyway I have never used a 80% acidified Methanol. But what I usually use acidified methanol (acidified with HCl) is for dissolving coelenterazine. What I use is 200 uL of 3N HCl in 9.8 ml of Methanol (every 10 ml of acidified Methanol contains 0.2 ml of 3N HCl).
More
VOTE
Wasim H Chowdhury
Dr.Wasim H Chowdhury
I am a student in analytical chemistry
and I want to know about preparing methanolic acid
I see your previously answer, and I hope to help me, Are there any references about this preparation
Thank you very much.
Wasim H Chowdhury
Dr.Wasim H Chowdhury
I am a student in analytical chemistry
and I want to know about preparing methanolic acid
I see your previously answer, and I hope to help me, Are there any references about this preparation
Thank you very much.
More
VOTE
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
More
VOTE
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
It depends on what you want. If you really want to have just HCl and CH3COH then you must either buy a gaseous HCl bottle and bubble it over MeOH, or produce HCl in your own lab and once again bubble it over CH3COH (it is an easy setup).
More
VOTE
The answer is unclear because the information you have provided is unclear. 1:80:10::HCl:MeOH:water v/v is clear although not a standard description. It means simply to combine 1mL HCl (which if not specified means conc. HCl) with 80mL of methanol and 10 mL of water. This however is not an 80% solution. What is probably meant is 1:80:20 v/v where the total exceeds 100% but that is a common way to describe a system when one of the components is much less than the others, i.e. 1:79.5:19.5 or some variation on this theme is very hard to process in the lab. The respondent talking about H2SO4 is missing the point - you don't need dry acidified methanol when 20% of the solution is water.
The answer is unclear because the information you have provided is unclear. 1:80:10::HCl:MeOH:water v/v is clear although not a standard description. It means simply to combine 1mL HCl (which if not specified means conc. HCl) with 80mL of methanol and 10 mL of water. This however is not an 80% solution. What is probably meant is 1:80:20 v/v where the total exceeds 100% but that is a common way to describe a system when one of the components is much less than the others, i.e. 1:79.5:19.5 or some variation on this theme is very hard to process in the lab. The respondent talking about H2SO4 is missing the point - you don't need dry acidified methanol when 20% of the solution is water.
More
VOTE