Chemical And Ionic Equilibrium

Chemical And Ionic Equilibrium Result 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 $\mathrm{NH} _4 \mathrm{OH}=4.7$ ]

(2019 Main, 9 Jan I)

(a) $9.3$

(b) $5.0$

(d) $5.2$

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Answer:

Correct Answer: 7. ( a )

Solution:

The reaction takes place when $\mathrm{H}_2 \mathrm{SO}_4$ is added to $\mathrm{NH}_4 \mathrm{OH}$ is as follows :

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,

$\begin{aligned} \mathrm{pOH} & =\mathrm{p} K_b+\log \frac{\left[\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]} \\ & =4.7+\log \frac{2}{2}=4.7 \\ \Rightarrow \quad \mathrm{pH} & =14-\mathrm{pOH}=14-4.7=9.3 \end{aligned}$