Question

$20mL$ of $0.1M$ $H_{2}S{O}_4$ solution is mixed with $30mL$ of $0.2M$ $N{H}_{4}OH$ solution. The $pH$ of the resultant solution is:
[$p{K}_b$ of $N{H}_{4}OH=4.7$]

• A. 7
• B. 9
• C. 5
• D. 12

Answer 1

The correct option is B 9

$mmol$ of $H_{2}S{O}_4=20\times 0.1=2$
$mmol$ of $N{H}_{4}OH=30\times 0.2=6$
$H_{2}S{O}_4+2N{H}_{4}OH\to (N{H}_4{)}_{2}S{O}_4+2H_{2}O$
$\begin{array}{ccc}\left(\text{Initial mmol}\right)& 2& 60\\ \left(\text{Final mmol}\right)& (2-2)& (6-2\times 2)2\end{array}$
$=0=2$
Since $H_{2}S{O}_4$ is the limiting reagent, it will be consumed completely.
In the mixture, finally we are left with $N{H}_{4}OH$ and $(N{H}_4{)}_{2}S{O}_4$, i.e., base and its salt with strong acid.
Such type of solutions are known as Basic Buffers.
$[N{H}_{4}OH{]}_{left}=2mmol[\left(N{H}_4{)}_{2}S{O}_4\right]=2mmol$
Here , one ($N{H}_4{)}_{2}S{O}_4$ salt gives two $N{H}_4^{+}$ ions.
$\therefore \left[N{H}_4^{+}\right]=2\times 2=4mmol$
Total volume $=30+20=50mL$
$pOH=p{K}_b+log\left[\frac{\text{Cation of Salt}}{\text{Base}}\right]$
$pOH=4.7+log\left(\frac{4/50}{2/50}\right)$
$pOH=4.7+log2=5$
$pH=14-pOH$
$pH=14-5=9$