Question

Calcium hydroxide, Ca(OH)2, is an ionic compound with a solubility product constant, Ksp, of 6.5×10–6. Calculate the solubility of this compound in pure water.

Answer 1

Answer: The solubility of this compound in pure water is 0.012 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)`

The equation for the ionization of the is given as:

`Ca(OH)_2\rightarrow Ca^(2+)+2OH^-`

By stoichiometry of the reaction:

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

When the solubility of
`Ca(OH)_2` is S moles/liter, then the solubility of
`Ca^(2+)` will be S moles\liter and solubility of
`OH^-` will be 2S moles/liter.

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

`6.5* 10^(-6)=[S][2S]^2`

`S=0.012M`

Thus solubility of this compound in pure water is 0.012 M