Question

(i) What is the pH of ${10}^{-8}NHCl$?
(ii) Calculate pH of the basic solutions having $\left[O{H}^{-}\right]$ as-
(a) $\left[O{H}^{-}\right]=0.05M$
(b) $\left[O{H}^{-}\right]=4\times {10}^{-6}M$

Answer 1

Since $H^{+}$ concentration from acid and water are comparable.

$[H^{+}{]}_{total}=[H^{+}{]}_{HCl}+[H^{+}{]}_{H_{2}O}$

Let $x$ amount of $H_{2}O$ dissociates and cocentration of $\left[H^{+}\right]fromionizationof$H_2O$isequalto$OH^-$ also.

$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$({10}^{-8}+x)\left(x\right)={10}^{-14}$

$x^2+{10}^{-8}x-{10}^{-14}$

$x^2+{10}^{-8}x-{10}^{-14}=0$

$x=9.5\times {10}^{-8}$

Therefore, $\left[H^{+}\right]=10.5\times {10}^{-8}=1.05\times {10}^{-7}$

$pH=-log\left[H^{+}\right]=6.98$

a) $pOH=-logO{H}^{-}=1.3$

$pH=14-1.3=12.7$

b) $pOH=6–2log2=5.4$

$pH=14-5.4=8.6$