At 25^{circ}C,the solubility product (K_{sp}) | vedclass

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(B) The dissociation of $NH_4OH$ is given by $NH_4OH \rightleftharpoons NH_4^+ + OH^-$.Given concentration of $NH_4OH = 0.1 \ M$ and degree of dissoci

Vedclass · Accepted Answer

$4 	imes 10^{-12}$

Vedclass · Answer

(B) The dissociation of $NH_4OH$ is given by $NH_4OH ightleftharpoons NH_4^+ + OH^-$.Given concentration of $NH_4OH = 0.1 \ M$ and degree of dissociation $\alpha = 50\% = 0.5$.Therefore,$[OH^-] = C 	imes \alpha = 0.1 	imes 0.5 = 0.05 \ M$.The solubility product expression for $Zn(OH)_2$ is $K_{sp} = [Zn^{2+}][OH^-]^2$.Substituting the values: $10^{-14} = [Zn^{2+}](0.05)^2$.$[Zn^{2+}] = \frac{10^{-14}}{(0.05)^2} = \frac{10^{-14}}{25 	imes 10^{-4}} = \frac{1}{25} 	imes 10^{-10} = 0.04 	imes 10^{-10} = 4 	imes 10^{-12} \ M$.

Salt	$K_{sp}$
$AgCl$	$2 \times 10^{-10}$
$AgBr$	$5 \times 10^{-13}$
$AgI$	$9 \times 10^{-17}$

At $25^{\circ}C$,the solubility product $(K_{sp})$ of $Zn(OH)_2$ is $10^{-14}$. If $NH_4OH$ is $50\%$ dissociated,what will be the concentration of zinc in a $0.1 \ M$ $NH_4OH$ solution?

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