Chemical and Ionic Equilibrium 2 Question 7
7. $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}{2} \mathrm{SO}{4}$ solution is added to $30 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{NH}{4} \mathrm{OH}$ solution. The $\mathrm{pH}$ of the resultant mixture is $\left[\mathrm{p} K{b}\right.$ of $\left.\mathrm{NH}_{4} \mathrm{OH}=4.7\right]$
(2019 Main, 9 Jan I)
(a) 9.3
(b) 5.0
(c) 9.0
(d) 5.2
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Answer:
Correct Answer: 7. (d)
Solution:
- The reaction takes place when $\mathrm{H}{2} \mathrm{SO}{4}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ is as follows :
Strong acid Weak base Salt of strong acid
Millimoles at $t=0 \quad 20 \times 0.1=2 \quad 30 \times 0.2=6 \quad \quad+$ weak base
Millimoles at $t=t \quad 0 \quad 2$
So, the resulting solution is a basic buffer $\left[\mathrm{NH}{4} \mathrm{OH}+\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4}\right]$.
According to the Henderson’s equation,
$$ \begin{aligned} \mathrm{pOH} & =\mathrm{p} K_{b}+\log \frac{\left[\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4}\right]}{\left[\mathrm{NH}{4} \mathrm{OH}\right]} \ & =4.7+\log \frac{2}{2}=4.7 \ \Rightarrow \quad \mathrm{pH} & =14-\mathrm{pOH}=14-4.7=9.3 \end{aligned} $$
