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.

Explanation:

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]

5.90×10-4=x2÷0.395-x

Explanation:

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]

pOH=-log⁡(x)=-log[OH]

pOH=-log[1.5266×10-2]

pOH=1.816

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

pH=14-pOH

& 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.

Leave a Reply

Your email address will not be published. Required fields are marked *

Call Now Button