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The concentration of chromate ions $ (CrO_4^{2-})$ at $15^oC$ in a saturated solution of pure strontium chromate is equal to $0.006\frac{mol}{L}$.
The equation for the dissolution of pure strontium chromate
$SrCrO_4(s)$ 
$Sr^{2+}(aq)$ + $CrO_4^{2-}(aq)$
shows that then the concentration of $Sr^{2+}$ is equal to .......... $\frac{mol}{L}$
So, the solubility product at $20^oC$ of $[SrCrO_4]$ is $K_s=[Sr^{2+}][CrO_4^{2-}]$ = .......... $\frac{mol^2}{L^2} $
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