Calculate the solublity of `SrSO_(4)` in water in `mol*L^(-1)` at `25^(@)C`. (Solublity product of `SrSO_(4)` at `25^(@)C=7.6xx10^(-7)mol^(2)*L^(-2)`]

Explanation

In saturated solution, the following equilibrium is set up by `SrSO_(4):SrSO_(4)(s)hArr Sr^(2+)(aq)+SO_(4)^(2-)(aq)`
Hence, solublity product of `SrSO_(4),(K_(sp))=[Sr^(2+)]xx[SO_(4)^(2-)]`
In in the solution, solubility of `SrSO_(4)` be S `mol*L^(-1)` then `[Sr^(2+)]=[SO_(4)^(2-)]=S" "mol*L^(-1)`
`thereforeK_(sp)=SxxS=S^(2)`. as per givenn data, `K_(sp)=7.6xx10^(-7)`
`therefore` At `25^(@)C`, solublity off `SrSO_(4)=8.7xx10^(-4)mol*L^(-1)`.