Question

There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc acetate, Zn(CH3COO)2, is 43.0 g/L, what is the solubility product Ksp of this compound?

Answer 1

Answer:

$K_{sp}$ of $Zn(CH_{3}COO)_{2}$ is 0.0513

Explanation:

Solubility equilibrium of $Zn(CH_{3}COO)_{2}$:

$Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}$

Solubility product of $Zn(CH_{3}COO)_{2}$ ($K_{sp}$) is written as-            $K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}$

Where $[Zn^{2+}]$ and $[CH_{3}COO^{-}]$ represents equilibrium concentration (in molarity) of $Zn^{2+}$ and $CH_{3}COO^{-}$ respectively.

Molar mass of $Zn(CH_{3}COO)_{2}$ = 183.48 g/mol

So, solubility of $Zn(CH_{3}COO)_{2}$ = $\frac{43.0}{183.48}M$ = 0.234M

1 mol of $Zn(CH_{3}COO)_{2}$ gives 1 mol of $Zn^{2+}$ and 2 moles of $CH_{3}COO^{-}$ upon dissociation.

so,   $[Zn^{2+}]$ = 0.234 M and $[CH_{3}COO^{-}]$ = $(2\times 0.234)M=0.468M$

so, $K_{sp}=(0.234)\times (0.468)^{2}=0.0513$