Question

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?

Answer 1

Answer:

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

Explanation:

$\text{From the information given:}$

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

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

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

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

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

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

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

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

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

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