The aqueous solution of which of the following salt will have $O H^{-}$ ions?

N a C l

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N a_{2} S O_{4}

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C H_{3} C O O N a

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None of these

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Solution

The correct option is D $C H_{3} C O O N a$
The aqueous solution of $N a C l$ and ${N a}_{2} {S O}_{4}$ will be neutral because both salts are formed from strong acid and strong base. $N a C l$ is formed from $N a O H$ and $H C l$ and ${N a}_{2} {S O}_{4}$ is formed $N a O H$ and $H_{2} {S O}_{4}$ . But ${C H}_{3} C O O N a$ is formed from weak acid ${C H}_{3} C O O H$ and strong base $N a O H$ . So hydrolysis of ${C H}_{3} C O O N a$ has been taken place to give ${O H}^{(-)}$ .
${C H}_{3} {C O O}^{(-)} + {N a}^{(+)} + H_{2} O = {C H}_{3} C O O H + {N a}^{+} + {O H}^{(-)}$ .
$∴$ So answer will be $C$ .

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The pH of 0.1M aqueous solutions of NaCl, $C H_{3}$ COONa and $N H_{4}$ Cl will follow the order

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N a_{2} S O_{4} + H_{2} S O_{4}

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