How many grams of Pb(IO3)2 (Ksp = 3.7 ✕ 10−13) will dissolve in 500. mL of a 0.0057 M PbCl2 solution? (Ksp = 1.7 ✕ 10−5)
 g

To solve this problem, we need to find how much lead(II) iodate, \( \text{Pb(IO}_3)_2 \), will dissolve in a solution of lead(II) chloride, \( \text{PbCl}_2 \), considering their solubility products (\( K_{sp} \)).

### Step 1: Define the equilibrium expression for \( \text{Pb(IO}_3)_2 \).

The dissolution of \( \text{Pb(IO}_3)_2 \) can be represented as:

\[
\text{Pb(IO}_3)_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{IO}_3^- (aq)
\]

The solubility product (\( K_{sp} \)) expression for this dissociation is:

\[
K_{sp} = [\text{Pb}^{2+}] [\text{IO}_3^-]^2
\]

We are given that:

\[
K_{sp} (\text{Pb(IO}_3)_2) = 3.7 \times 10^{-13}
\]

### Step 2: Account for the common ion effect.

The solution already contains lead ions (\( \text{Pb}^{2+} \)) from the dissociation of \( \text{PbCl}_2 \). The dissociation of \( \text{PbCl}_2 \) is:

\[
\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq)
\]

Since the solution has 0.0057 M \( \text{PbCl}_2 \), the concentration of \( \text{Pb}^{2+} \) from the \( \text{PbCl}_2 \) will be:

\[
[\text{Pb}^{2+}] = 0.0057 \, \text{M}
\]

This lead ion concentration from \( \text{PbCl}_2 \) reduces the solubility of \( \text{Pb(IO}_3)_2 \) due to the common ion effect.

### Step 3: Set up the \( K_{sp} \) expression.

Let \( s \) be the molar solubility of \( \text{Pb(IO}_3)_2 \) in moles per liter. Therefore, the concentration of iodate ions \( [\text{IO}_3^-] \) will be \( 2s \), and the concentration of lead ions \( [\text{Pb}^{2+}] \) from \( \text{Pb(IO}_3)_2 \) will add a negligible amount to the 0.0057 M from the \( \text{PbCl}_2 \) solution, so we approximate:

\[
[\text{Pb}^{2+}] \approx 0.0057 \, \text{M}
\]

Substituting into the \( K_{sp} \) expression:

\[
K_{sp} = [\text{Pb}^{2+}] [\text{IO}_3^-]^2
\]

\[
3.7 \times 10^{-13} = (0.0057)(2s)^2
\]

### Step 4: Solve for \( s \).

Simplifying the equation:

\[
3.7 \times 10^{-13} = (0.0057)(4s^2)
\]

\[
3.7 \times 10^{-13} = 0.0228s^2
\]

\[
s^2 = \frac{3.7 \times 10^{-13}}{0.0228}
\]

\[
s^2 = 1.6228 \times 10^{-11}
\]

\[
s = \sqrt{1.6228 \times 10^{-11}}
\]

\[
s = 1.2737 \times 10^{-6} \, \text{M}
\]

### Step 5: Convert molar solubility to grams.

The molar mass of \( \text{Pb(IO}_3)_2 \) is:

\[
\text{Pb(IO}_3)_2 = 1 \times 207.2 + 2 \times (126.9 + 3 \times 16) = 207.2 + 2 \times 174.9 = 557 \, \text{g/mol}
\]

The solubility in moles per liter is \( 1.2737 \times 10^{-6} \, \text{mol/L} \). In 500 mL (0.5 L), the amount of \( \text{Pb(IO}_3)_2 \) that dissolves is:

\[
\text{moles} = 1.2737 \times 10^{-6} \times 0.5 = 6.3685 \times 10^{-7} \, \text{mol}
\]

Finally, the mass of \( \text{Pb(IO}_3)_2 \) that dissolves is:

\[
\text{mass} = 6.3685 \times 10^{-7} \, \text{mol} \times 557 \, \text{g/mol} = 3.5475 \times 10^{-4} \, \text{g} = 0.00035475 \, \text{g}
\]

### Final Answer:

Approximately **0.00035 grams** of \( \text{Pb(IO}_3)_2 \) will dissolve in 500 mL of a 0.0057 M \( \text{PbCl}_2 \) solution.

---

Would you like more details or have any questions? Here are 5 related questions to expand your understanding:

1. How does the common ion effect influence solubility?
2. What is the role of \( K_{sp} \) in determining solubility?
3. Why was the \( [\text{Pb}^{2+}] \) concentration approximated as 0.0057 M?
4. How would the solubility change if the concentration of \( \text{PbCl}_2 \) were higher?
5. What is the significance of molar solubility in chemical equilibria?

**Tip:** When working with solubility product constants, always account for the common ion effect, as it reduces the solubility of salts.

How many grams of Pb(IO3)2 (Ksp = 3.7 ✕ 10−13) will dissolve in 500. mL of a 0.0057 M PbCl2 solution? (Ksp = 1.7 ✕ 10−5)

Find out how to calculate the solubility of Pb(IO3)2 in a 0.0057 M PbCl2 solution. This detailed solution uses Ksp and accounts for the common ion effect, providing step-by-step instructions for advanced chemistry students. Learn how to apply solubility equilibrium principles to solve this problem, suitable for Grades 11-12.

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