This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, [tex]9.33\times 10^{-5}mol/L[/tex]
Explanation :
As we know that CaCO₃ dissociates to give [tex]Ca^{2+}[/tex] ion and [tex]CO_3^{2-}[/tex] ion.
The solubility equilibrium reaction will be:
[tex]CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ca^{2+}][CO_3^{2-}][/tex]
Let solubility of CaCO₃ be, 's'
[tex]K_{sp}=(s)\times (s)[/tex]
[tex]K_{sp}=s^2[/tex]
[tex]8.7\times 10^{-9}=s^2[/tex]
[tex]s=9.33\times 10^{-5}mol/L[/tex]
Therefore, the solubility of CaCO₃ is, [tex]9.33\times 10^{-5}mol/L[/tex]