Which of the following represents the correct order of the acidity for equimolar aqueous solution for $H C l$ , $H_{2} S O_{4}$ , $N H_{4} O H$ and $N a O H$ ?

H C l < N H_{4} O H < N a O H < H_{2} S O_{4}

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N H_{4} O H < N a O H < H_{2} S O_{4} < H C l

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H C l < H_{2} S O_{4} < N H_{4} O H < N a O H

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N a O H < N H_{4} O H < H C l < H_{2} S O_{4}

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Solution

The correct option is D $N a O H < N H_{4} O H < H C l < H_{2} S O_{4}$
$N a O H < N H_{4} O H < H C l < H_{2} S O_{4}$ represents the increasing order of the acid strength.
$N a O H$ is a strong base. Ammonium hydroxide is a weak base.

$H C l$ is a strong acid and sulphuric acid is the strongest acid.

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F e (O H)_{3}

A l (O H)_{3}

\begin{matrix} Column - I & Column - II (Enthalpy change in kcal) & (Neutralisation) (A) & < 13.7 kcal & (p) & H C l (1 m o l) + N a O H (1 m o l) (B) & = 13.7 kcal & (q) & H F (1 m o l) + N a O H (1 m o l) (C) & > 13.7 kcal & (r) & N H_{4} O H (1 m o l) + H C l (1 m o l) (D) & = 27.4 kcal & (s) & N a O H (2 m o l) + H_{2} S O_{4} (1 m o l) (t) & N a O H (1 m o l) + C H_{3} C O O H (1 m o l) \end{matrix}

N H_{4}

C H_{3}

C H_{3}

N H_{4}

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