Q13 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Chemistry
The $\mathrm{pH}$ of the solution, when
(i) sodium acetate is dissolved in water.
(ii) ammonium chloride is dissolved in water.
- Correct answerA(i) $\mathrm{pH}>7$ (ii) $\mathrm{pH} < 7$
- B(i) $\mathrm{pH}=7$ (ii) $\mathrm{pH}=0$
- C(i) $\mathrm{pH}>7$ (ii) $\mathrm{pH}>7$
- D(i) $\mathrm{pH} < 7$ (ii) $\mathrm{pH}>7$
Solution
$\because$ (i) Sodium acetate, i.e. $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ in water gives $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{NaOH}$ as follows :
$\mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}$
Hence, for $(\mathrm{i}) \mathrm{pH}>7$. $\quad[\because$ solution is basic $]$
Also,
(ii) when, ammonium chloride, i.e. $\left(\mathrm{NH}_4 \mathrm{Cl}\right)$ is added in water, it gives $\mathrm{NH}_4^{+}$(aqueous) and $\mathrm{Cl}^{-}$ (aqueous) ions.
$\mathrm{NH}_4^{+} \text {(aqueous) }+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_3+\mathrm{H}^{+}+\mathrm{OH}^{-}$
Hence, $\mathrm{pH} < 7$
Hence, the correct answer is, for (i) $\mathrm{pH}>7$ and for (ii) $\mathrm{pH} < 7$.
Hence, the correct answer is option: 'A'.