The ${K_{sp}}$ of $BaSO_4$ is $1.1 \times 10^{-10}$. Will a precipitate form when equal volumes of $2 \times 10^{-4} \ M \ BaCl_2$ and $5.0 \times 10^{-3} \ M \ H_2SO_4$ solutions are mixed? Explain by calculation.

(A) When equal volumes are mixed,the total volume doubles,so the concentration of each reactant is halved:
$[Ba^{2+}] = \frac{2 \times 10^{-4}}{2} = 1.0 \times 10^{-4} \ M$
$[SO_4^{2-}] = \frac{5.0 \times 10^{-3}}{2} = 2.5 \times 10^{-3} \ M$
The ionic product $Q_{sp}$ is calculated as:
$Q_{sp} = [Ba^{2+}][SO_4^{2-}] = (1.0 \times 10^{-4}) \times (2.5 \times 10^{-3}) = 2.5 \times 10^{-7}$
Since $Q_{sp} (2.5 \times 10^{-7}) > K_{sp} (1.1 \times 10^{-10})$,a precipitate of $BaSO_4$ will be formed.