Question

What is the molar solubility of mn(oh)2(s) in a solution that is buffered at ph 8.00 at 25 °c? the ksp of mn(oh)2 is 1.9 ´ 10–13 at 25 °c?

Answer 1

Answer: The molar solubility of
`Mn(OH)_2` in a solution that is buffered at ph 8.00 is 0.19 M

Step-by-step explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as
`K_(sp)`

We are given:

Solubility product of
`Mn(OH)_2` =
`1.9* 10^(-13)`

The equation for the ionization of the
`Mn(OH)_2` is given as:

`Mn(OH)_2\leftrightharpoons Mn^(2+)+2OH^(-)`

1 mole of
`Mn(OH)_2` gives 1 mole of
`Mn^(2+)` and 2 moles of
`OH^(-)`

Given : pH = 8.00

`pH+pOH=14`

`pOH=14.0-8.00=6.00`

`6.00=-log[OH^-]`

`[OH^-]=10^(-6)M`

`K_(sp)=[Mn^(2+)][OH^(-)]^2`

`1.9* 10^(-13)M=[s][(10^(-6))^2]`

`s=0.19M`

Thus the molar solubility of
`Mn(OH)_2` in a solution that is buffered at ph 8.00 is 0.19 M