`Pb(IO_(3))_(2)` is a sparingly soluble salt `(K_(sp) = 2.6 xx 10^(-13))`. To `35mL` of `0.15M Pb(NO_(3))_(2)` solution, `15mL` of `0.8M KIO_(3)` solution is added, and a precipiatte of `Pb(IO_(3))_(2)` is formed.
What will be molarity of `Pb^(2+)` ions in the solution after completion of the reactions? `[IO3^(-)]=0.03M`

Options

A. `8.4 xx 10^(-10)`
B. `1.6 xx 10^(-10)`
C. `2.8 xx 10^(-10)`
D. `6.1 xx 10^(-10)`

Explanation

To find the molarity of \( \text{Pb}^{2+} \) ions in the solution after the precipitation of \( \text{Pb(IO}_3)_2 \), we can follow these steps:

### Step 1: Calculate the moles of \( \text{Pb}^{2+} \) and \( \text{IO}_3^{-} \)

1. **Calculate moles of \( \text{Pb}^{2+} \)**:
\[ \text{Volume of } \text{Pb(NO}_3)_2 = 35 \, \text{mL} = 0.035 \, \text{L} \]
\[ \text{Concentration of } \text{Pb(NO}_3)_2 = 0.15 \, \text{M} \]
\[ \text{Moles of } \text{Pb}^{2+} = \text{Concentration} \times \text{Volume} = 0.15 \, \text{mol/L} \times 0.035 \, \text{L} = 0.00525 \, \text{mol} \]

2. **Calculate moles of \( \text{IO}_3^{-} \)**:
\[ \text{Volume of } \text{KIO}_3 = 15 \, \text{mL} = 0.015 \, \text{L} \]
\[ \text{Concentration of } \text{KIO}_3 = 0.8 \, \text{M} \]
\[ \text{Moles of } \text{IO}_3^{-} = \text{Concentration} \times \text{Volume} = 0.8 \, \text{mol/L} \times 0.015 \, \text{L} = 0.012 \, \text{mol} \]

### Step 2: Determine the limiting reactant

The reaction for the formation of \( \text{Pb(IO}_3)_2 \) is:
\[ \text{Pb}^{2+} + 2 \text{IO}_3^{-} \rightarrow \text{Pb(IO}_3)_2 \]

From the stoichiometry of the reaction, 1 mole of \( \text{Pb}^{2+} \) reacts with 2 moles of \( \text{IO}_3^{-} \).

- Moles of \( \text{IO}_3^{-} \) required for \( 0.00525 \, \text{mol} \) of \( \text{Pb}^{2+} \):
\[ 2 \times 0.00525 = 0.0105 \, \text{mol} \]

Since we have \( 0.012 \, \text{mol} \) of \( \text{IO}_3^{-} \), \( \text{Pb}^{2+} \) is the limiting reactant.

### Step 3: Calculate remaining moles of \( \text{IO}_3^{-} \)

After the reaction:
- Moles of \( \text{IO}_3^{-} \) that react:
\[ 0.00525 \, \text{mol} \times 2 = 0.0105 \, \text{mol} \]

- Remaining moles of \( \text{IO}_3^{-} \):
\[ 0.012 \, \text{mol} - 0.0105 \, \text{mol} = 0.0015 \, \text{mol} \]

### Step 4: Calculate the total volume of the solution

Total volume after mixing:
\[ 35 \, \text{mL} + 15 \, \text{mL} = 50 \, \text{mL} = 0.050 \, \text{L} \]

### Step 5: Calculate the concentration of \( \text{Pb}^{2+} \) ions in the solution

Using the \( K_{sp} \) expression for \( \text{Pb(IO}_3)_2 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{IO}_3^{-}]^2 \]
Given \( K_{sp} = 2.6 \times 10^{-13} \) and \( [\text{IO}_3^{-}] = 0.03 \, \text{M} \):
\[ 2.6 \times 10^{-13} = [\text{Pb}^{2+}](0.03)^2 \]
\[ 2.6 \times 10^{-13} = [\text{Pb}^{2+}](0.0009) \]
\[ [\text{Pb}^{2+}] = \frac{2.6 \times 10^{-13}}{0.0009} = 2.89 \times 10^{-10} \, \text{M} \]

### Final Answer

The molarity of \( \text{Pb}^{2+} \) ions in the solution after completion of the reactions is approximately \( 2.89 \times 10^{-10} \, \text{M} \).

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