The solubility of `Sr(OH)_(2)` at `298 K` is `19.23 g L^(-1)` of solution. Calculate the concentrations cf strontium and hydroxyl ions and the `pH` of the solution.

Answer

13

Explanation

Molar mass of `Sr(OH)_2` = 87.6 + 2 x 17 = 121.6
Solubilty = `19.23/121.6`
=0.1581 M
`Sr(OH)_2` ionizes as :
`Sr(OH)_2=Sr^(2+)+2OH^(-)`
`[Sr^(2+)]`=0.1581 M, `[OH^-]` = 2 x 0.1581 = 0.3162 M
Now, `[H_3O^+][OH^-]=K_w`
`[H_3O^+]=(1xx10^(-14))/([OH^-])`
`=(1xx10^(-14))/0.3162=3.163xx10^(-14)`
pH=-log `[H_2O^+]`
=-log `(3.163xx10^(-14))`
=-(0.499-14)
=13.50.