An aqueous solution contains an unknown concentration of Ba$$^{2+}$$. When 50 mL of a 1 M solution of Na$$_2$$SO$$_4$$ is added, BaSO$$_4$$ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO$$_4$$ is $$1 \times 10^{-10}$$. What is the original concentration of Ba$$^{2+}$$?

We are told that a saturated barium sulphate system is reached the moment a known amount of sodium sulphate is introduced. At that instant the ionic product of the two ions equals the solubility product.

First, we recall the definition of the solubility product. For the equilibrium

$$\mathrm{BaSO_4(s)} \rightleftharpoons \mathrm{Ba^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}$$

the solubility product is written as

$$K_{sp}= [\mathrm{Ba^{2+}}]\,[\mathrm{SO_4^{2-}}].$$

Numerically, the problem states that $$K_{sp}=1 \times 10^{-10}.$$

Now we calculate the concentration of the sulphate ion immediately after the sodium sulphate solution is mixed in, but before any significant precipitation occurs.

Moles of sulphate supplied:

$$n_{\mathrm{SO_4^{2-}}}=0.050\ \text{L}\times 1\ \text{mol L}^{-1}=0.050\ \text{mol}.$$

The final volume of the mixture is given as $$0.500\ \text{L}$$ (500 mL). Hence the sulphate ion concentration right at the start of precipitation is

$$[\mathrm{SO_4^{2-}}]=\frac{0.050\ \text{mol}}{0.500\ \text{L}}=0.10\ \text{M}.$$

At the point where precipitation just begins we must have

$$[\mathrm{Ba^{2+}}]_{\text{mixed}}\,[\mathrm{SO_4^{2-}}]=K_{sp}.$$

Substituting $$K_{sp}=1\times10^{-10}$$ and $$[\mathrm{SO_4^{2-}}]=0.10\ \text{M},$$ we obtain

$$[\mathrm{Ba^{2+}}]_{\text{mixed}}\times0.10=1\times10^{-10}.$$

So,

$$[\mathrm{Ba^{2+}}]_{\text{mixed}}=\frac{1\times10^{-10}}{0.10}=1\times10^{-9}\ \text{M}.$$

This value is the barium ion concentration in the solution after it has been diluted to the final 500 mL. The question, however, asks for the concentration in the original 450 mL of the unknown solution.

Let the initial concentration be $$C_0\ (\text{mol L}^{-1}).$$ Dilution obeys the relation

$$C_0 \times V_0 = C_{\text{diluted}} \times V_{\text{final}},$$

where $$V_0=0.450\ \text{L}$$ and $$V_{\text{final}}=0.500\ \text{L}.$$ Rearranging gives

$$C_0=\frac{C_{\text{diluted}}\times V_{\text{final}}}{V_0}.$$

Substituting the known numbers,

$$C_0=\frac{(1\times10^{-9}\ \text{M})\times0.500\ \text{L}}{0.450\ \text{L}} =\frac{5.0\times10^{-10}}{0.450} =1.11\times10^{-9}\ \text{M}.$$

Rounding to two significant figures gives $$1.1\times10^{-9}\ \text{M}.$$

Hence, the correct answer is Option D.