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Arrange the aqueous solutions of the following salts in the increasing order of $\mathrm{pH}$
$$
\begin{array}{ccc}
\mathrm{CuSO}_4 & \mathrm{NaCN} & \mathrm{KCl} \\
\mathrm{I} & \text { II } & \text { III }
\end{array}
$$
Options:
$$
\begin{array}{ccc}
\mathrm{CuSO}_4 & \mathrm{NaCN} & \mathrm{KCl} \\
\mathrm{I} & \text { II } & \text { III }
\end{array}
$$
Solution:
2779 Upvotes
Verified Answer
The correct answer is:
I < III < II
$\mathrm{NaCN}$ is the salt of strong base $(\mathrm{NaOH})$ and weak acid (HCN). So, its $\mathrm{pH}>7$.
$\mathrm{CuSO}_4$ is a salt of strong acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$ and a weak base $\left[\mathrm{Cu}(\mathrm{OH})_2\right.$, thus its solution is acidic with $\mathrm{pH} < 7$.
$\mathrm{KCl}$ is a salt of strong acid $(\mathrm{HCl})$ and a strong base $[\mathrm{KOH}]$, thus its solution is neutral with $\mathrm{pH}=7$.
Increasing order of $\mathrm{pH}$ is
$$
\begin{array}{ccc}
\mathrm{CuSO}_4 & < \mathrm{KCl} & < \mathrm{NaCN} \\
\text { I } & \text { III } & \text { II }
\end{array}
$$
$\mathrm{CuSO}_4$ is a salt of strong acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$ and a weak base $\left[\mathrm{Cu}(\mathrm{OH})_2\right.$, thus its solution is acidic with $\mathrm{pH} < 7$.
$\mathrm{KCl}$ is a salt of strong acid $(\mathrm{HCl})$ and a strong base $[\mathrm{KOH}]$, thus its solution is neutral with $\mathrm{pH}=7$.
Increasing order of $\mathrm{pH}$ is
$$
\begin{array}{ccc}
\mathrm{CuSO}_4 & < \mathrm{KCl} & < \mathrm{NaCN} \\
\text { I } & \text { III } & \text { II }
\end{array}
$$
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