[lukas.stib]

Hello chemists,

I have several esters of carbox. acids at home and I want to test the basic hydrolysis of esters with NaOH. I can calculate the stoichiometry of reactants and products, the problem is I don't know how much water is used to dissolve NaOH? Is there any principle how much water to use or does it not matter?

Thanks, Lukáš S.

[AWK]

Check solubility of NaOH in water.
But your problem is rather to get a homogeneous mixture or emulsion to carry out the hydrolysis reactions relatively quickly.

[lukas.stib]

OK, thank you. I always use a slight excess of NaOH. It doesn't have to be very fast, I always reflux for an hour.

[lukas.stib]

Solubility in water is 1000 g / l 25°C. So, for example, if I have to use 30 g of NaOH in stoichiometry, I will I dissolve it in any amount? Or 30 g H2O?

For homogeneous mixture or emulsion I can used excess water.
 

[AWK]

The hydrolysis of esters from NaOH is practically quantitative and is not disturbed by alcohol. Methanol and ethanol dissolve NaOH quite well. So if you are hydrolyzing methyl or ethyl esters, it is better to use NaOH solutions in alcohol. In this way, you will obtain insoluble sodium salts of acids and you have solved the problem of homogeneity of the solution during the reaction. Always vigorous stirring speeds up the reaction.
A pretty good approximation is 1g NaOH per 1ml of water, 5ml methanol, or 10ml ethanol.

[lukas.stib]

So, if I understand correctly, when hydrolyzing methyl salicylate (for example), I dissolve NaOH in methanol (according to the approximation you gave) and pour into methyl salicylate?

When researching theories, they always wrote only an aqueous solution of NaOH.

Super for your approximation, I have to remeber her.

[AWK]

NaOH always contains some sodium carbonate. In addition, some sodium carbonate is formed when the reaction mixture is stirred, unless the reaction is carried out under an inert gas atmosphere. Therefore, a slight excess of NaOH, several percent, must be used for the hydrolysis. The sodium salts of some organic acids are somewhat soluble in alcohol. This excess of NaOH also significantly reduces their solubility. After separation of the salts, and before recovering the alcohol by distillation, sodium carbonate and excess NaOH must be neutralized with sulfuric acid, and the precipitate filtered off.

There is no need to use absolute alcohol.

[lukas.stib]

OK, thank you very much!!! ;-)