Question

The Ksp for Zn(OH)2 is 5.0 x 10-17. Determine the molar solubility of Zn(OH)2 in a buffer solution with a pH of 11.5.
a) 5.0 x 106
b) 1.2 x 10-12
c) 1.6 x 10-14
d) 5.0 x 10-12
e) 5.0 x 10-17

Answer 1

A is the answer
5.0x106

Answer 2

Answer : The correct option is, (d) $5.0\times 10^{-12}mole$

Explanation :

First we have to calculate the $H^+$ concentration.

$pH=-\log [H^+]$

$11.5=-\log [H^+]$

$[H^+]=3.16\times 10^{-12}M$

Now we have to calculate the $OH^-$ concentration.

$[H^+][OH^-]=K_w$

$3.16\times 10^{-12}\times [OH^-]=1.0\times 10^{-14}$

$[OH^-]=3.16\times 10^{-3}M$

Now we have to calculate the molar solubility of $Zn(OH)_2$.

The balanced equilibrium reaction will be:

$Zn(OH)_2\rightleftharpoons Zn^{2+}+2OH^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Zn^{2+}][OH^-]^2$

Now put all the given values in this expression, we get:

$5.0\times 10^{-17}=[Zn^{2+}]\times (3.16\times 10^{-3})^2$

$[Zn^{2+}]=5.0\times 10^{-12}M$

Therefore, the molar solubility of $Zn(OH)_2$ is, $5.0\times 10^{-12}mole$