Question

Good day

When Zn(OH)2(s) was added to 1.00 L of a basic solution, 1.15x 10^-2 mol of the solid dissolved.
What is the concentration of OH^- in the final solution?
Some info:
-The solubility-product constant for Zn(OH)2 is Ksp= 3.00 x 10^-16
-The formation constant for the hydroxo complex, Zn(OH)4 2- , is Kf= 4.60 x 10^17

Thank you so much.

Answer 1

Solution.

$Zn(OH)2 = Zn^{2+} + 2OH^-$

$Ksp = [Zn^{2+}] \times [OH^-]^2$

$Zn(OH)2 + 2OH^- = [Zn(OH)4]^{2-}$

$n(Zn(OH)2) = n([Zn(OH)4]^{2-})$

$[Zn(OH)4]^{2-} = \frac{1.15 \times 10^{-2}}{1} = 1.15 \times 10^{-2} M$

$Kf = \frac{[Zn(OH)4]^{2-}}{[Zn^{2+}] \times [OH^-]^4}$

$Kf = \frac{[Zn(OH)4]^{2-}}{Ksp(Zn(OH)2) \times [OH^-]^2}$

$[OH^-] = (\frac{[Zn(OH)4]^{2-}}{Kf \times Ksp(Zn(OH)2)})^{0.5}$

$[OH^-] = 9.13 \times 10^{-3} M$

Answer:

$[OH^-] = 9.13 \times 10^{-3} M$