An aqueous solution contains an unknown concentration of Ba$^{2+}$. When 50 mL of a 1 M Na$_2$SO$_4$ solution is added, BaSO$_4$ just begins to precipitate. The final volume is 500 mL and $K_{sp}(\text{BaSO}_4) = 1 \times 10^{-10}$. What was the original concentration of Ba$^{2+}$?
At the point of precipitation: $[\text{Ba}^{2+}][\text{SO}_4^{2-}] = K_{sp} = 1 \times 10^{-10}$
Find $[\text{SO}_4^{2-}]$ in the final mixture:
Moles of SO$_4^{2-}$ added $= 0.050\ \text{L} \times 1\ \text{M} = 0.05\ \text{mol}$
Final volume $= 500\ \text{mL} = 0.5\ \text{L}$
$[\text{SO}_4^{2-}]_{\text{final}} = \dfrac{0.05}{0.5} = 0.1\ \text{M}$
Find $[\text{Ba}^{2+}]$ in the final mixture:
$[\text{Ba}^{2+}]_{\text{final}} = \dfrac{K_{sp}}{[\text{SO}_4^{2-}]} = \dfrac{1\times10^{-10}}{0.1} = 1\times10^{-9}\ \text{M}$
Back-calculate original concentration: Final volume is 10× original Ba$^{2+}$ volume (500 mL total, 450 mL original solution).
Original $[\text{Ba}^{2+}] = 1\times10^{-9} \times \dfrac{500}{450} \approx \mathbf{2\times10^{-9}}\ \text{M}$
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