The solubility of a salt of weak acid (AB) at | vedclass

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(A) The solubility $(S)$ of a salt of a weak acid $AB$ in an acidic medium is given by the formula: $S = \sqrt{K_{sp} \left(1 + \frac{[H^{+}]}{K_{a}}\

Vedclass · Accepted Answer

$4.47$

Vedclass · Answer

(A) The solubility $(S)$ of a salt of a weak acid $AB$ in an acidic medium is given by the formula: $S = \sqrt{K_{sp} \left(1 + \frac{[H^{+}]}{K_{a}}ight)}$Given: $K_{sp} = 2 	imes 10^{-10}$,$K_{a} = 1 	imes 10^{-8}$,and $pH = 3$,so $[H^{+}] = 10^{-3} \ M$.Substituting the values:$S = \sqrt{2 	imes 10^{-10} \left(1 + \frac{10^{-3}}{10^{-8}}ight)}$$S = \sqrt{2 	imes 10^{-10} \left(1 + 10^{5}ight)}$Since $10^{5} \gg 1$,we can approximate $1 + 10^{5} \approx 10^{5}$.$S = \sqrt{2 	imes 10^{-10} 	imes 10^{5}} = \sqrt{2 	imes 10^{-5}}$$S = \sqrt{20 	imes 10^{-6}} = \sqrt{20} 	imes 10^{-3}$Comparing this with $Y 	imes 10^{-3}$,we get $Y = \sqrt{20} \approx 4.47$.

The solubility of a salt of weak acid $(AB)$ at $pH = 3$ is $Y \times 10^{-3} \ mol \ L^{-1}$. The value of $Y$ is (Given that the value of solubility product of $AB$ $(K_{sp}) = 2 \times 10^{-10}$ and the value of ionization constant of $HB$ $(K_{a}) = 1 \times 10^{-8}$)

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