Question

Calculate the amount of ammonium chloride required to dissolve in $500mL$ water to have $pH=4.5$ ($K_b$ for ${NH}_{4}OH$ is $1.8\times {10}^{-5}$)

Answer 1

$\left[H^{+}\right]={10}^{-pH}={10}^{-5}=3.162\times {10}^{-5}M$

Let $C$ be the concentration of ${NH}_{4}Cl$

${NH}_4^{+}+H_{2}O\rightleftharpoons {NH}_{4}OH+H^{+}$

$C(1-h)$ $Ch$ $Ch$

If $h$ is small then

$K_h=C{h}^2[K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}]$

$h=\frac{K_h}{Ch}=\frac{K_h}{\left[H^{+}\right]}=\frac{5.5\times {10}^{-10}}{3.162\times {10}^{-5}}=1.739\times {10}^{-5}$

$C=\frac{\left[H^{+}\right]}{h}=\frac{3.162\times {10}^{-5}}{1.739\times {10}^{-5}}=1.8mol{L}^{-1}$

$500mL$ of water contains$=\frac{1.8}{2}=0.9mol$

Mass in $g=0.9\times 53.5=48.15g$