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

Answer : The correct option is, (d)
`5.0* 10^(-12)mole`

Explanation :

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

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

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

`[H^+]=3.16* 10^(-12)M`

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

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

`3.16* 10^(-12)* [OH^-]=1.0* 10^(-14)`

`[OH^-]=3.16* 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* 10^(-17)=[Zn^(2+)]* (3.16* 10^(-3))^2`

`[Zn^(2+)]=5.0* 10^(-12)M`

Therefore, the molar solubility of
`Zn(OH)_2` is,
`5.0* 10^(-12)mole`