An aqueous solution contains an unknown concentration of Ba^ 2 + ....

$\left[Ba^{2 +}\right] =1{0}^{- 9}$ M (in 500 mL solution)

\left[S{O}_{4}^{2 -}\right]$in 500 mL solution will be<br /><br />$50 \times 1 = M \times 500$<br /><br />M$= 0.1$<br /><br />$\mathop{Ba^{2 +}}_{450\text{mL}} + \mathop{N{a}_{2}S{O}_{4}}_{50\text{mL, 1M}} \to BaS{O}_{4} + 2Na^{+}$<br /><br />${K}_{sp} \left(BaS{O}_{4}\right) = \left[Ba^{2 +}\right]\left[S{O}_{4}^{2 -}\right]$<br /><br />$1{0}^{- 10} = \left[Ba^{2 +}\right]\times0.1$<br /><br />Now, we have to calculate \left[Ba^{2 +}\right]$ in original solution (450 mL)

$1{0}^{- 9} \times500 = 450 \timesM$

$M = \dfrac{1{0}^{- 9} \times 500}{450} = \dfrac{10}{9} \times1{0}^{- 9} = 1.11 \times1{0}^{- 9}M$