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 K_{b} and for weak base dissociation is less in solution.

### Step 2 of 2

K_{b}=5.9×10^{-4} for the weak base dimethylamine.

K_{b}=Dissociation constant.

+

CH_{3})_{2}NH + H2O ↽−−⇀ (CH_{3})_{2}NH_{2} + OH^{–}

Initial concentration 0.395 0 0

Concentration at equilibrium 0.395-x x x

+

K_{b}=[ (CH_{3})_{2}NH_{2}] [OH^{–}]÷[(CH_{3})_{2}NH]

5.90×10^{-4}=x^{2}÷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}=x^{2}÷0.395

So x^{2}= 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 K_{b }value.Then after from pOH value we can calculate the pH value.