Question

Solution of $0.1N{{H}_{4}}OH$ and $0.1N\,N{{H}_{4}}Cl$ has pH 9.25. Then, $p{{K}_{b}}$ of $N{{H}_{4}}OH$ is

• A. 9.25
• B. 4.75
• C. 3.75
• D. 8.25

Answer 1

Answer: (B)

$pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]}$ $\because$ $[N{{H}_{4}}Cl]=[N{{H}_{4}}OH]$ $\therefore$ $pOH=p{{K}_{b}}$ $p{{K}_{b}}=pOH=14-pH$ $=14-9.25=4.75$