What is the molar solubility of Mn(OH)_2 (K_{ | vedclass

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(B) For a basic buffer,the Henderson-Hasselbalch equation is: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$.Since the moles of $NH_4^+$ and $NH_3$ are e

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$1.38 	imes 10^{-4}$

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(B) For a basic buffer,the Henderson-Hasselbalch equation is: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$.Since the moles of $NH_4^+$ and $NH_3$ are equal,$\frac{[NH_4^+]}{[NH_3]} = 1$,so $pOH = pK_b = -\log(1.8 	imes 10^{-5}) \approx 4.74$.Therefore,$[OH^-] = K_b = 1.8 	imes 10^{-5} \ M$.The solubility product expression for $Mn(OH)_2$ is $K_{sp} = [Mn^{2+}][OH^-]^2 = S 	imes [OH^-]^2$.Substituting the values: $4.5 	imes 10^{-14} = S 	imes (1.8 	imes 10^{-5})^2$.$S = \frac{4.5 	imes 10^{-14}}{3.24 	imes 10^{-10}} = 1.388 	imes 10^{-4} \ M \approx 1.38 	imes 10^{-4} \ M$.

What is the molar solubility of $Mn(OH)_2$ $(K_{sp} = 4.5 \times 10^{-14})$ in a buffer solution containing equal moles of $NH_4^+$ and $NH_3$ $(K_b = 1.8 \times 10^{-5})$?

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