The solubility of `Sr(OH)_(2)` at 298K is 19.23 g/L of solution. Calculate the concentrations of stronium and hydroxyl ions and the pH of the solution.

Explanation

`Sr(OH)_(2)(s)hArr Sr^(2+)(aq)+2OH^(-)(aq)`
Given: Solubility of `Sr(OH)_(2)` is `19.23g*L^(-1)`
`19.23g*L^(-1)Sr(OH)_(2)-=(19.23)/((87.62+34))mol*L^(-1)`
`-=0.1581mol*L^(-1)Sr(OH)_(2)`
[Molar mass of `Sr(OH)_(2)=(87.62+34)g*mol^(-1)`]
`therefore` In this solution of `Sr(OH)_(2),[Sr^(2+)]=0.1581mol*L^(-1)`
`[OH^(-)]=2xx0.1581=0.3162mol*L^(-1)`
For this solution, `pOH=-log_(10)[OH^(-)]`
`=-log_(10)(0.3162)=0.5`
and `pH=14-pOH=14-0.5=13.5`