Question

How many grams of ammonium chloride should be dissolved in 500 mL of water to have a solution of pH 4.5?

[$K_b$ for ammonium hydroxide is $1.8\times {10}^{-5}$]

• A. 25.22
• B. 48.15
• C. 53.5
• D. 90

Answer 1

The correct option is B 48.15

$\left[H^{+}\right]={10}^{-pH}={10}^{-4.5}=3.162\times {10}^{-5}mol/L$
$\left[H^{+}\right]=\sqrt{\frac{K_{w}C}{K_b}}$
$C=\frac{[H^{+}{]}^2{K}_b}{K_w}$
$C=\frac{3.162\times {10}^{-5}\times 3.162\times {10}^{-5}\times 1.8\times {10}^{-5}}{1\times {10}^{-14}}=1.8mol/L$
Number of moles of ammonium chloride present in 500 mL $=\frac{1.8\times 500}{1000}=0.9$ moles.
The molar mass of ammonium chloride is 53.5 g/mol. The mass of ammonium chloride will be $0.90\times 53.5=48.15$ g