Let's begin by finding the solubility of lead (II) chloride $$\mathrm{PbCl}_2$$ in water from its solubility product constant $$(\mathrm{K}_{\mathrm{SP}})$$.

The dissolution of lead (II) chloride can be represented by the following equation:

$$\mathrm{PbCl}_2(s) \rightarrow \mathrm{Pb}^{2+}(aq) + 2\mathrm{Cl}^{-}(aq)$$

For the equilibrium:

$$\mathrm{K}_{\mathrm{SP}} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2$$

Let $$s$$ be the solubility of $$\mathrm{PbCl}_2$$ in mol/L. From the dissociation equation, it can be seen that the concentration of $$\mathrm{Pb}^{2+}$$ ions in solution is $$s$$ and the concentration of $$\mathrm{Cl}^{-}$$ ions is $$2s$$. Insert these into the solubility product expression:

$$\mathrm{K}_{\mathrm{SP}} = (s)(2s)^2 = 4s^3$$

Given that $$\mathrm{K}_{\mathrm{sp}} = 3.2 \times 10^{-8}$$, we can solve for $$s$$:

$$4s^3 = 3.2 \times 10^{-8}$$

$$s^3 = \frac{3.2 \times 10^{-8}}{4}$$

$$s^3 = 8 \times 10^{-9}$$

$$s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \mathrm{~mol/L}$$

This is the molar solubility of $$\mathrm{PbCl}_2$$. Next, calculate the amount of water required to dissolve $$0.2 \mathrm{~g}$$ of $$\mathrm{PbCl}_2$$.

The molar mass of $$\mathrm{PbCl}_2$$ is given as $$278 \mathrm{~g/mol}$$. To find the number of moles of $$0.2 \mathrm{~g}$$ of $$\mathrm{PbCl}_2$$:

$$n = \frac{mass}{molar\ mass} = \frac{0.2 \mathrm{~g}}{278 \mathrm{~g/mol}} \approx 7.194 \times 10^{-4} \mathrm{~mol}$$

Since the solubility of $$\mathrm{PbCl}_2$$ is $$2 \times 10^{-3} \mathrm{~mol/L}$$ (from the previous calculation), the volume of water $$V$$ needed to dissolve these moles can be calculated by:

$$V = \frac{n}{s} = \frac{7.194 \times 10^{-4} \mathrm{~mol}}{2 \times 10^{-3} \mathrm{~mol/L}} = 0.3597 \mathrm{~L}$$

Convert this to milliliters:

$$0.3597 \mathrm{~L} = 359.7 \mathrm{~mL}$$

Thus, the volume of water needed to dissolve $$0.2 \mathrm{~g}$$ of $$\mathrm{PbCl}_2$$ is approximately $$359.7 \mathrm{~mL}$$.

So, the correct answer is Option B - 359.7 mL.

Ionic Equilibrum

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