The pH of 0.1 M solution of the following sa | vedclass

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(B) $HCl$ is a strong acid. In its $0.1 \ M$ solution,$[H^{+}] = 0.1 \ M$,hence $pH = 1$.$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $

Vedclass · Accepted Answer

$HCl < NH_4Cl < NaCl < NaCN$

Vedclass · Answer

(B) $HCl$ is a strong acid. In its $0.1 \ M$ solution,$[H^{+}] = 0.1 \ M$,hence $pH = 1$.$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis to form an acidic solution $(pH $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis,resulting in a neutral solution $(pH = 7)$.$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis to form an alkaline solution $(pH > 7)$.Therefore,the $pH$ increases in the order: $HCl < NH_4Cl < NaCl < NaCN$.

The $pH$ of $0.1 \ M$ solution of the following salts increases in the order

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