Chemical and Ionic Equilibrium 2 Question 7
7.
(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}
Strong acid Weak base Salt of strong acid
Millimoles at
Millimoles at
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,
$$ \begin{aligned} \mathrm{pOH} & =\mathrm{p} K_{b}+\log \frac{\left[\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4}\right]}{\left[\mathrm{NH}{4} \mathrm{OH}\right]} \ & =4.7+\log \frac{2}{2}=4.7 \ \Rightarrow \quad \mathrm{pH} & =14-\mathrm{pOH}=14-4.7=9.3 \end{aligned} $$
