Question

A $0.01M$ ammonia solution is $5$% ionised, its $pH$ will be:

• A. $11.80$
• B. $10.69$
• C. $7.22$
• D. $12.24$

Answer 1

Answer: (B)

$10.69$

(i) $\left[O{H}^{-}\right]=C\propto$ for weak acids/ bases.
(ii) $pOH=-\log \left[O{H}^{-}\right]$$\left(iii\right)$pH = 14 - pOH$Given,aqueoussolutionof$NH_{3}$\left(weakbase\right)$C = 0.01\ M\alpha = 5= \dfrac {5}{100}[OH^{-}] = C\alpha = 0.01\times \dfrac {5}{100} = 5\times 10^{-4}pOH = -\log [OH^{-}]= -\log (5\times 10^{-4})= 4\log 10 - \log 5= 4 - 0.6989= 3.3010\therefore pH = 14 - pOH = 14 - 3.3010 = 10.6990$