Question

$20mL$ of $0.1M{H}_{2}S{O}_4$ solution is added to $30mL$ of $0.2MN{H}_{4}OH$ solution. The $pH$ of the resultant mixture is $[p{K}_{b}ofN{H}_{4}OH=4.7,\log 2=0.3010]$

• A. 5.2
• B. 9
• C. 5
• D. 9.4

Answer 1

The correct option is B 9

$\text{Millimoles of}{H}_{2}S{O}_4=20\times 0.1=2$
$\text{Millimoles of}N{H}_{4}OH=30\times 0.2=6$

$H_{2}S{O}_4+2N{H}_{4}OH\to (N{H}_4{)}_{2}S{O}_4+2H_{2}O$
$Initial260$
$Equilibrium06-4=22$

$\text{Total volume}=30+20=50mL$
$\left[\right(N{H}_4{)}_{2}S{O}_4]=\frac{2}{50}M\therefore [N{H}_4^{+}]=\frac{2}{50}\times 2=\frac{4}{50}M[N{H}_{4}OH]=\frac{2}{50}M$

$pOH=p{K}_b+\log \frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}=4.7+\log \frac{\frac{4}{50}}{\frac{2}{50}}pOH=4.7+\log 2\approx 5.0\therefore pH=14-pOHpH=14-5=9.0$