Question

A 0.01m ammonia solution is 5% ionised its ph will be

Answer 1

Answer : The pH will be, 10.7

Solution :  Given,

Concentration (C) = 0.01 M

Degree of dissociation $(\alpha)$ = 5% = 0.05

The equilibrium reaction for dissociation of ammonia base is,

                            $NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

initially conc.         c                       0         0

At eqm.              $c(1-\alpha)$                $c\alpha$        $c\alpha$

First we have to calculate the concentration of $[OH^-]$

$[OH^-]=c\alpha$

$[OH^-]=(0.01)\times (0.05)=5\times 10^{-4}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (5\times 10^{-4})$

$pOH=3.3$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-3.3=10.7$

Therefore, the pH will be, 10.7