1. Condition for just beginning of precipitation

When $BaSO_4$ just begins to precipitate,

$[Ba^{2+}]_{\text{final}}[SO_4^{2-}]_{\text{final}} = K_{sp} = 1\times 10^{-10}$

2. Find sulfate concentration after mixing

Given:

• Volume of $Na_2SO_4$ added $= 50\,\text{mL} = 0.050\,\text{L}$
• Concentration of $Na_2SO_4 = 1\,\text{M}$

Moles of $SO_4^{2-}$ added:

$n_{SO_4^{2-}} = 1\times 0.050 = 0.050\,\text{mol}$

Final volume $= 500\,\text{mL} = 0.500\,\text{L}$

So final sulfate concentration is

$[SO_4^{2-}]_{\text{final}} = \frac{0.050}{0.500} = 0.10\,\text{M}$

3. Find final $Ba^{2+}$ concentration at precipitation point

Using $K_{sp}$:

$[Ba^{2+}]_{\text{final}} = \frac{K_{sp}}{[SO_4^{2-}]_{\text{final}}} = \frac{1\times 10^{-10}}{0.10} = 1\times 10^{-9}\,\text{M}$

4. Relate final concentration to original concentration

Let original concentration of $Ba^{2+}$ be $C$.

Before adding $Na_2SO_4$, let the original volume of the $Ba^{2+}$ solution be:

$500 - 50 = 450\,\text{mL} = 0.450\,\text{L}$

Moles of $Ba^{2+}$ initially present:

$n_{Ba^{2+}} = C\times 0.450$

After dilution to $0.500\,\text{L}$, just before precipitation begins,

$[Ba^{2+}]_{\text{final}} = \frac{C\times 0.450}{0.500} = 0.9C$

But we found:

$0.9C = 1\times 10^{-9}$

So,

$C = \frac{1\times 10^{-9}}{0.9} = 1.11\times 10^{-9}\,\text{M}$

5. Match with options

$C \approx 1.1\times 10^{-9}\,\text{M}$

So the correct option is D.