A monoprotic acid in a 0.1 M solution ionize | vedclass

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(D) For a weak monoprotic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H^{+} + A^{-}$.Given concentration $C = 0.1 \ M$ and degree o

Vedclass · Accepted Answer

$1.0 	imes 10^{-11}$

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(D) For a weak monoprotic acid $HA$,the dissociation is given by: $HA ightleftharpoons H^{+} + A^{-}$.Given concentration $C = 0.1 \ M$ and degree of ionization $\alpha = 0.001 \% = \frac{0.001}{100} = 10^{-5}$.The ionization constant $K_a$ is given by the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.Since $\alpha$ is very small $(10^{-5} \ll 1)$,we can approximate $1 - \alpha \approx 1$.Thus,$K_a \approx C \alpha^2 = 0.1 	imes (10^{-5})^2 = 10^{-1} 	imes 10^{-10} = 10^{-11}$.

$A$ monoprotic acid in a $0.1 \ M$ solution ionizes to $0.001 \%$. Its ionisation constant is

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