Question

The pH of 0.1 M aqueous solution of $NaCl,C{H}_{3}COONa$ and $N{H}_{4}Cl$ will follow the order __________.

• A. $NaCl>N{H}_{4}Cl$
• B. $N{H}_{4}Cl>NaCl$
• C. $N{H}_{4}Cl=NaCl$
• D. None of these

Answer 1

Answer: (B)

$N{H}_{4}Cl>NaCl$

The salts will ionize as:

$NaCl\rightleftharpoons {Na}^{+}+{Cl}^{-}C{H}_{3}COONa\rightleftharpoons {Na}^{+}+{C{H}_{3}COO}^{-}N{H}_{4}Cl\rightleftharpoons {N{H}_4}^{+}+{Cl}^{-}$

$NaCl$ is formed from strong acid and strong base therefore both will ionize completely and will result in neutral pH=7

$C{H}_{3}COONa$ is a salt of strong base and weak acid, therefore will have completely dissociated base and partially dissociated acid, which results in basic solution with pH>7

$N{H}_{4}Cl$ is a salt of weak base and strong acid. Upon ionization, it gives partially ionized base and completely ionized acid, thereby making the solution acidic with pH<7.

Therefore pH will follow the order:

$N{H}_{4}Cl$<$NaCl$<$C{H}_{3}COONa$