Answer : The value of [tex]K_{sp}[/tex] for barium sulfate is, [tex]1\times 10^{-10}M^2[/tex]
Solution :
The balanced equilibrium reaction will be,
[tex]BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ba^{2+}][SO_4^{2-}][/tex]
Let the concentration of barium ion and sulfate ion be, 's'.
[tex]K_{sp}=(s)\times (s)[/tex]
[tex]K_{sp}=(s)^2[/tex]
From the balanced equilibrium reaction, we conclude that the concentration of sulfate will be equal to the concentration of barium.
Now put all the value of in the above expression, we get the value of solubility constant for barium sulfate.
[tex]K_{sp}=(1\times 10^{-5})^2[/tex]
[tex]K_{sp}=1\times 10^{-10}M^2[/tex]
Therefore, the value of [tex]K_{sp}[/tex] for barium sulfate is, [tex]1\times 10^{-10}M^2[/tex]
