Answer: The solubility of calcium sulfate is [tex]8.43\times 10^{-3}M[/tex]
Explanation:
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of calcium sulfate follows:
[tex]CaSO_4(aq.)\rightleftharpoons Ca^{2+}(aq.)+SO_4^{2-}(aq.)[/tex]
s s
The expression of [tex]K_{sp}[/tex] for above equation follows:
[tex]K_{sp}=s\times s[/tex]
We are given:
[tex]K_{sp}=7.10\times 10^{-5}[/tex]
Putting values in above expression, we get:
[tex]7.10\times 10^{-5}=s\times s\\\\s=\sqrt{7.10\times 10^{-5}}=8.43\times 10^{-3}M[/tex]
Hence, the solubility of calcium sulfate is [tex]8.43\times 10^{-3}M[/tex]
