The POH of an aqueous solution of 0.395M dimethylamine ( a weak base with the formula (CH3)2NH is

Screenshot 2024 01 19 232426

Screenshot 2024 01 19 232426

poh of aqueous solution

Step 1 of 2

Given is the 0.395M solution of weak base of dimethylamine.

We have found out the pOH of given solution with concentration of 0.395M.


Here we use the concept of equilibrium constant Kb and for weak base dissociation is less in solution.

Step 2 of 2

Kb=5.9×10-4 for the weak base dimethylamine.

Kb=Dissociation constant.


CH3)2NH + H2O ↽−−⇀ (CH3)2NH2 + OH

Initial concentration 0.395 0 0

Concentration at equilibrium 0.395-x x x


Kb=[ (CH3)2NH2] [OH]÷[(CH3)2NH]



We use this equilibrium constant formula and ignore the value of x in denominator for term 0.395-x as this is a weak base so dissociate less in solution.

So, we can write this as 5.90×10-4=x2÷0.395

So x2= 5.90×10-4 ×0.395

and x= (5.90×10-4 ×0.395)1/2

x=1.5266 ×10-2=[OH]




Final solution

Hence the pOH of given solution is 1.816=1.82 after rounding off.

Note for finding out the pH we use the formula, pH+pOH=14


& pOH=14-pH

But for the base we calculate the pOH from its Kb value.Then after from pOH value we can calculate the pH value.

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