Answer:
The precipitate will form.
Explanation:
Let's write the equilibrium expression for the solubility product of calcium sulfate:
[tex]CaSO_4(s)[/tex] ⇄ [tex]Ca^{2+}(aq)+SO_4^{2-}(aq)[/tex]
The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:
[tex]K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}[/tex]
Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:
[tex]Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}[/tex]
Now given the K value of:
[tex]K_{sp}=10^{-4.5}=3.2\cdot10^{-5}[/tex]
Notice that:
[tex]Q_{sp}>K_{sp}[/tex]
This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.
