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vlabodo [156]
2 years ago
13

The solubility product for Zn(OH)2 is 3.0×10−16. The formation constant for the hydroxo complex, Zn(OH)2−4, is 4.6×1017. What is

the minimum concentration of OH− required to dissolve 1.7×10−2 mol of Zn(OH)2 in a liter of solution?
Chemistry
1 answer:
Yuri [45]2 years ago
8 0

Answer:

[OH^-]= 1.12 \times 10^{-2} \ M

Explanation:

\text{From the information given:}

K_{sp} \ of \ Zn(OH)_2 \ is \ 3.0 \times 10^{-16}

K_f \ of \ Zn(OH)_2 = 4.6 \times 10^{17}

\mathtt{K_{sp} \ \ \ \ \ \ \ \ \mathtt{Zn(OH)_{2(s)}} \to Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)} }

\mathtt{K_{f} \ \ \ \ \ \ \ \ \mathtt{Zn^{2+}_{(aq)} } + 4OH^-_{(aq)} \to Zn(OH)^{2-}_{4}_{(aq)}}

\mathtt{K \ \ \ \ \ \ \ \ \mathtt{{Zn(OH)_{2}_{(s)}}} + 2OH^{-}_{(aq)}  \to Zn(OH)^{2-} _{4(aq) }}

\text{To calculate the minimum concentration of OH}^{-}

K = \dfrac{Zn(OH)^-_4}{[OH^-]^2} \\ \\

[OH^-]^2= \dfrac {0.017}{1.380 \times 10^2}

[OH^-]^2=1.232 \times 10^{-4}

[OH^-]= 1.12 \times 10^{-2} \ M

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