The heats of neutralisation of $C H_{3} C O O H$ , HCOOH, HCN and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

$H_{2} S < H C N < H C O O H < C H_{3} C O O H$

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$H C N < H_{2} S < C H_{3} C O O H < H C O O H$

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$H C O O H < C H_{3} C O O H < H_{2} S < H C N$

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$C H_{3} - C O O H < H C O O H < H C N < H_{2} S$

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Solution

The correct option is B
$H C N < H_{2} S < C H_{3} C O O H < H C O O H$

Heat of neutralization per equivalent $\propto$ Acidic strength, the maximum value being the enthalpy of formation of water. Every equivalent of an acid would form 1 mole of water if it were a strong acid. As an acid gets weaker, this value decreases as fewer water molecules are formed - as fewer protons are furnished by weaker acids.

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and

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are –12.8, –13.2, –3.2 and –3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

The heats of neutralisation of $C H_{3} C O O H$ , HCOOH, HCN and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

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and

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under identical conditions is:

(

K_{a}

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and

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)

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The heats of neutralisation of CH3COOH, HCOOH, HCN and H2S are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

The heats of neutralisation of $C H_{3} C O O H$ , HCOOH, HCN and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.