Question

A $20mL$ solution of $0.5MN{H}_{4}OH$ is to be titrated with $0.2MHCl$. Find the $pH$ of solution initially i.e. when no $HCl$ is added.
($K_b\left(N{H}_{4}OH\right)=1.8\times {10}^{-5}$)

• A. 11.48
• B. 2.52
• C. 10.48
• D. 3.52

Answer 1

The correct option is A 11.48

mmol of $N{H}_{4}OH=20\times 0.5=10$
mmol of $HCl=0$
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)10000$
Since only weak base $N{H}_{4}OH$ is present initially.
$pO{H}_{initial}=-log\sqrt{K_{b}C}pO{H}_{initial}=-log\sqrt{1.8\times {10}^{-5}\times 0.5}$
$pO{H}_{initial}=\frac{1}{2}\left(log\right(9\times {10}^{-6}\left)\right)pO{H}_{initial}=\frac{1}{2}(6-0.96)=\frac{5.04}{2}pO{H}_{initial}=2.52$
$p{H}_{inital}=14-2.52=11.48$