Answer
Yes, there will be $CaSO_4$ precipitation.
Work Step by Step
1. Calculate the molar mass $(Na_2SO_4)$:
22.99* 2 + 32.07* 1 + 16* 4 ) = 142.05g/mol
2. Calculate the number of moles $(Na_2SO_4)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.105}{ 142.05}$
$n(moles) = 7.392\times 10^{- 4}$
3. Find the concentration in mol/L $(Na_2SO_4)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 7.392\times 10^{- 4}}{ 0.1000} $
$C(mol/L) = 7.392\times 10^{- 3}$
Therefore, the $[SO_4^{2-}]$ is equal to $7.392 \times 10^{-3}M$
4. Calculate the $Q_{sp}$:
- $K_{sp}$ expression: $[Ca^{2+}][SO_4^{2-}]$
$Q_{sp} = 0.025 \times 7.392 \times 10^{-3} = 1.8 \times 10^{-4}$
- The $K_{sp}$ for this salt is equal to $7.10 \times 10^{-5}$
Therefore, since the calculated value is greater, there will be precipitation.