Question

Prove that $pH=\frac{1}{2}[pKw-pKb-lo{g}_{10}C]$ where $h<5\%$.

Answer 1

In case of cation hydrolysis of salt of a weak base and strong acid,

$H_{2}O\rightleftharpoons H^{+}+{OH}^{-}{K}_w\underset{–}{{NH}_4^{+}+{OH}^{-}\rightleftharpoons {NH}_{4}OH\frac{1}{K_b}}\underset{–}{{NH}_4^{+}+H_{2}O⟶{NH}_{4}OH+H^{+}}{K}_h=\frac{K_w}{K_b}=K_a--------\left(1\right)$

${NH}_4^{+}+H_{2}O⟶{NH}_{4}OH+H^{+}att=0Cat{eqb}^{m}C-ChChCh$

where h is the degree of hydrolysis

$K_h=\frac{\left[{NH}_{4}OH\right]\left[H^{+}\right]}{\left[{NH}_4^{+}\right]}=\frac{C^2{h}^2}{C(1-h)}=\frac{C{h}^2}{1-h}$

As h<5%, we can neglect h and thus

$K_h=C{h}^2⟹h=\sqrt{\frac{K_h}{C}}$

From equation (1), $K_h=\frac{K_w}{K_b}$

$⟹h=\sqrt{\frac{K_w}{K_b\cdot C}}$

As $\left[H^{+}\right]=Ch=C\cdot \sqrt{\frac{K_w}{K_b\cdot C}}=\sqrt{\frac{K_w\cdot C}{K_b}}$

$pH=-\log \left[H^{+}\right]=-\frac{1}{2}[\log K_w+\log C-\log K_b]$

$pH=\frac{1}{2}[p{K}_w-p{K}_b-{\log }_{10}C]$