The volume of water required to dissolve 0.1 g PbCl\(_2\) to get a saturated solution (in mL) is (Given K\(_{sp}\)(PbCl\(_2\)) = \(3.2 \times 10^{-8}\); Atomic mass of Pb = 207u)
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For a sparingly soluble salt of type \(A_x B_y\), the relationship between \(K_{sp}\) and molar solubility (s) is \(K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}\). For PbCl\(_2\), it's type \(AB_2\) (x=1, y=2), which gives \(K_{sp} = 1^1 2^2 s^{1+2} = 4s^3\).
Step 1: Relate the solubility product constant (K\(_{sp}\)) to the molar solubility (s).
The dissolution of lead(II) chloride is: \( \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) \).
If the molar solubility of PbCl\(_2\) is 's' mol/L, then at equilibrium, [Pb\(^{2+}\)] = s and [Cl\(^-\)] = 2s.
The expression for K\(_{sp}\) is: \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \).
\( K_{sp} = (s)(2s)^2 = 4s^3 \).
Step 2: Calculate the molar solubility (s).
We are given \( K_{sp} = 3.2 \times 10^{-8} \).
\( 3.2 \times 10^{-8} = 4s^3 \).
\( s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} = 8 \times 10^{-9} \).
\( s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \) mol/L.
Step 3: Calculate the solubility in grams per liter.
Molar mass of PbCl\(_2\) = (Atomic mass of Pb) + 2 \(\times\) (Atomic mass of Cl).
Molar mass \(\approx 207 + 2 \times 35.5 = 207 + 71 = 278\) g/mol.
Solubility in g/L = Molar solubility (mol/L) \(\times\) Molar mass (g/mol).
Solubility = \( (2 \times 10^{-3}) \times 278 = 556 \times 10^{-3} = 0.556 \) g/L.
This means that 1 liter (1000 mL) of water can dissolve 0.556 g of PbCl\(_2\) to form a saturated solution.
Step 4: Calculate the volume of water needed to dissolve 0.1 g of PbCl\(_2\).
We can set up a proportion:
\( \frac{\text{Volume of water}}{\text{Mass dissolved}} = \frac{1000 \text{ mL}}{0.556 \text{ g}} \).
Volume required = \( \frac{1000 \text{ mL}}{0.556 \text{ g}} \times 0.1 \text{ g} \).
Volume required = \( \frac{100}{0.556} \approx 179.85 \) mL.
This value is approximately 180 mL.
