- Language
- 🇷🇺
- Joined
- Jun 3, 2023
- Messages
- 23
- Reaction score
- 38
- Points
- 13
And so, this post is designed more for new users. the purpose of this post is to get rid of small mistakes during the preparation of methylamine. if you already know everything and this article is not new to you, then please just put a bold "Like" and you can add your experience in the comments.
For the preparation of aqueous methylamine, as you probably already know, you will need directly methylamine hydrochloride, distilled water, alkali - aka sodium hydroxide NaOH or potassium hydroxide KOH.
Let's begin my dear reader. I will take calculations for 150 grams of methylamine hydrochloride hcl
Calculations to obtain a solution of methylamine from 150 grams of methylamine hydrochloride:
1. Dissolution of alkali:
2. Adding alkali to methylamine hydrochloride:
Do not add all of the alkali at once, as this may cause a sudden spike in pH and loss of methylamine.!!!!!!
It is recommended to add the alkali in small portions (20-30 mL each), stirring after each addition.
Monitor the pH of the solution:
Use a pH meter to monitor the pH of the solution.
Required pH value:
For NaOH: 11-12
For KOH: 12-13
Add alkali until the target pH value is reached. I consider this step to be the most important when preparing aqueous methylamine.
100 g of pure alkali (NaOH or KOH) should be dissolved in distilled water.
Volume of water:
About 250-300 ml of distilled water will be required to completely dissolve 100 g of alkali.
The exact volume of water may vary slightly depending on the alkali used and the desired concentration of the solution.
Stirring:
In the desired container ( glass jar in my case), pour in the lye and pour in water.
Stir the solution thoroughly until the lye is completely dissolved.
You can use a magnetic stirrer or just shake the flask by hand.
REMEMBER AND BE CAREFUL: THE LYE WILL HEAT UP WHEN IT COMES INTO CONTACT WITH WATER. You can prepare a container with cold water in advance by adding ice to it. in case of strong heating, cool down in cold water - this is from my experience. After complete dissolution of the lye in distilled water, leave the jar to cool to room temperature, as it cools down go to the second stage.
2. Adding alkali to methylamine hydrochloride:
Do not add all of the alkali at once, as this may cause a sudden spike in pH and loss of methylamine.!!!!!!
It is recommended to add the alkali in small portions (20-30 mL each), stirring after each addition.
Monitor the pH of the solution:
Use a pH meter to monitor the pH of the solution.
Required pH value:
For NaOH: 11-12
For KOH: 12-13
Add alkali until the target pH value is reached. I consider this step to be the most important when preparing aqueous methylamine.
3. precipitation and filtration:
After all the amount of alkali has been added and the pH has reached the target value, allow the solution to settle.
A precipitate of sodium chloride (NaCl) will fall to the bottom.
Filter the solution through a filter (such as a vacuum filter).
Rinse the sodium chloride precipitate with a small amount of water to remove any residual methylamine - this is optional.
NARRATING.
150 g methylamine hydrochloride (CH₃NH₂HCl)
100 g of alkali (NaOH) of 98% purity
250 ml distilled water
1. Calculate the molar concentration of methylamine hydrochloride:
Molar mass of CH₃NH₂HCl = 74.53 g/mol
Moles of CH₃NH₂HCl = 150 g / 74.53 g/mol = 2.01 mol
2. Calculate the molar concentration of the alkali:
NaOH molar mass = 40 g/mol
NaOH purity = 98%
Moles of NaOH = (100 g * 0.98) / 40 g/mol = 2.45 mol
3. Determine the volume of the solution:
Volume of solution = 250 mL = 0.25 liters
4. Calculate the molar concentration of methylamine (CH₃NH₂):
In the neutralization reaction of methylamine hydrochloride (CH₃NH₂HCl) with an alkali (NaOH), methylamine (CH₃NH₂) and water (H₂O) are formed:
CH₃NH₂HCl + NaOH → CH₃NH₂ + NaCl + H₂O
From the reaction equation, it can be seen that 1 mole of CH₃NH₂HCl produces 1 mole of CH₃NH₂.
Therefore, the molar concentration of CH₃NH₂ in solution is:
[CH₃NH₂] = (2.01 mol CH₃NH₂HCl) / 0.25 L = 8.04 mol/L The calculations are probably not quite correct, but those who have more experience and knowledge, please take part in correcting the calculations.
That's all. Don't forget to subscribe and put a fat like if it was useful. Thank you.
P/S I would like to make a special mention and express my gratitude: @Marvin "Popcorn" Sutton and @WillD , and, of course, the forum: BBgate.com
For the preparation of aqueous methylamine, as you probably already know, you will need directly methylamine hydrochloride, distilled water, alkali - aka sodium hydroxide NaOH or potassium hydroxide KOH.
Let's begin my dear reader. I will take calculations for 150 grams of methylamine hydrochloride hcl
Calculations to obtain a solution of methylamine from 150 grams of methylamine hydrochloride:
1. Dissolution of alkali:
2. Adding alkali to methylamine hydrochloride:
Do not add all of the alkali at once, as this may cause a sudden spike in pH and loss of methylamine.!!!!!!
It is recommended to add the alkali in small portions (20-30 mL each), stirring after each addition.
Monitor the pH of the solution:
Use a pH meter to monitor the pH of the solution.
Required pH value:
For NaOH: 11-12
For KOH: 12-13
Add alkali until the target pH value is reached. I consider this step to be the most important when preparing aqueous methylamine.
100 g of pure alkali (NaOH or KOH) should be dissolved in distilled water.
Volume of water:
About 250-300 ml of distilled water will be required to completely dissolve 100 g of alkali.
The exact volume of water may vary slightly depending on the alkali used and the desired concentration of the solution.
Stirring:
In the desired container ( glass jar in my case), pour in the lye and pour in water.
Stir the solution thoroughly until the lye is completely dissolved.
You can use a magnetic stirrer or just shake the flask by hand.
REMEMBER AND BE CAREFUL: THE LYE WILL HEAT UP WHEN IT COMES INTO CONTACT WITH WATER. You can prepare a container with cold water in advance by adding ice to it. in case of strong heating, cool down in cold water - this is from my experience. After complete dissolution of the lye in distilled water, leave the jar to cool to room temperature, as it cools down go to the second stage.
2. Adding alkali to methylamine hydrochloride:
Do not add all of the alkali at once, as this may cause a sudden spike in pH and loss of methylamine.!!!!!!
It is recommended to add the alkali in small portions (20-30 mL each), stirring after each addition.
Monitor the pH of the solution:
Use a pH meter to monitor the pH of the solution.
Required pH value:
For NaOH: 11-12
For KOH: 12-13
Add alkali until the target pH value is reached. I consider this step to be the most important when preparing aqueous methylamine.
3. precipitation and filtration:
After all the amount of alkali has been added and the pH has reached the target value, allow the solution to settle.
A precipitate of sodium chloride (NaCl) will fall to the bottom.
Filter the solution through a filter (such as a vacuum filter).
Rinse the sodium chloride precipitate with a small amount of water to remove any residual methylamine - this is optional.
NARRATING.
150 g methylamine hydrochloride (CH₃NH₂HCl)
100 g of alkali (NaOH) of 98% purity
250 ml distilled water
1. Calculate the molar concentration of methylamine hydrochloride:
Molar mass of CH₃NH₂HCl = 74.53 g/mol
Moles of CH₃NH₂HCl = 150 g / 74.53 g/mol = 2.01 mol
2. Calculate the molar concentration of the alkali:
NaOH molar mass = 40 g/mol
NaOH purity = 98%
Moles of NaOH = (100 g * 0.98) / 40 g/mol = 2.45 mol
3. Determine the volume of the solution:
Volume of solution = 250 mL = 0.25 liters
4. Calculate the molar concentration of methylamine (CH₃NH₂):
In the neutralization reaction of methylamine hydrochloride (CH₃NH₂HCl) with an alkali (NaOH), methylamine (CH₃NH₂) and water (H₂O) are formed:
CH₃NH₂HCl + NaOH → CH₃NH₂ + NaCl + H₂O
From the reaction equation, it can be seen that 1 mole of CH₃NH₂HCl produces 1 mole of CH₃NH₂.
Therefore, the molar concentration of CH₃NH₂ in solution is:
[CH₃NH₂] = (2.01 mol CH₃NH₂HCl) / 0.25 L = 8.04 mol/L The calculations are probably not quite correct, but those who have more experience and knowledge, please take part in correcting the calculations.
That's all. Don't forget to subscribe and put a fat like if it was useful. Thank you.
P/S I would like to make a special mention and express my gratitude: @Marvin "Popcorn" Sutton and @WillD , and, of course, the forum: BBgate.com