Question

Calculate amount of $N{H}_{4}Cl$ (in $g$) required to be dissolved in $500mL$ of water to have $pH=4.5[K_{b}N{H}_{4}OH=1.8\times {10}^{-5}].$
(Given ${10}^{-4.5}=3.162\times {10}^{-5}$)

Answer 1

$\left[H^{+}\right]={10}^{-pH}={10}^{-4.5}=3.162\times {10}^{-5}M$
$\underset{C(1-h)}{N{H}_4^{+}+H_{2}O}\rightleftharpoons \underset{Ch}{N{H}_{4}OH}+\underset{Ch}{H^{+}}$
$K_h=C{h}^2=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$
$h=\frac{K_h}{Ch}=\frac{K_h}{\left[H^{+}\right]}\times \frac{5.5\times {10}^{-10}}{3.162\times {10}^{-5}}=1.74\times {10}^{-5}$
$C=\frac{\left[H^{+}\right]}{h}=\frac{3.162\times {10}^{-5}}{1.74\times {10}^{-5}}=1.8mol/L$
500 ml of $H_{2}O$ contains $\frac{1.8}{2}=0.9$ mole
Mass in g $=0.9\times 53.5=48.15g$