Question

Determine the pH of a $0.05 \,M$ aqueous solution of ammonium sulphate, $ k_{b} $ of ammonia is $ 2\times 10^{-5} $ ?

• A. 4.30
• B. 5.15
• C. 7.0
• D. 8.75

Answer 1

Answer: (B)

The hydrolysis reaction is
$ NH _{4}^{+}+ H _{2} O \rightleftharpoons NH _{4} OH + H ^{+} $
$ K _{ h }=\frac{\left[ NH _{4} OH \right]\left[ H ^{+}\right]}{\left[ NH _{4}\right]}=\frac{ K _{ w }}{ K _{5}}=5 \times 10^{-10} $
$5 \times 10^{-10}=\frac{\left[ H ^{+}\right]^{2}}{ c }\left\{\because\left[ NH _{4} OH \right]=\left[ H ^{-}\right]\right\}$
${\left[ H ^{+}\right]=\sqrt{5 \times 10^{-10} \times 2 \times 0.05}}$
${\left[ H ^{+}\right]=7 \times 10^{-6} \Rightarrow pH =5.15}$