In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains `Ag^(+) " and " Pb^(+)` at a concentration of 0.10M. Aqueous HCl is added to this solution until be `Cl^(-)` concentration is 0.10M. What will be concentration of `Ag^(+) " and " Pb^(2+)` be at equilibrium ?
(`K_(sp) " for AgCl " = 1.8xx10^(-10)`
`K_(sp) " for " PbCl_(2) = 1.7xx10^(-5)`)

Options

A. `[Ag^(+)]=1.8xx10^(-11)M,`
`[Pb^(2+)]=1.7xx10^(-4)M`
B. `[Ag^(+)]=1.8xx10^(-7)M,`
`[Pb^(2+)]=1.7xx10^(-6)M`
C. `[Ag^(+)]=1.8xx10^(-11)M,`
`[Pb^(2+)]=8.5xx10^(-5)M`
D. `[Ag^(+)]=1.8xx10^(-9)M,`
`[Pb^(2+)]=1.7xx10^(-3)M`

Explanation

To solve the problem, we need to determine the equilibrium concentrations of \( \text{Ag}^+ \) and \( \text{Pb}^{2+} \) ions after adding \( \text{HCl} \) to the solution until the concentration of \( \text{Cl}^- \) reaches 0.10 M. We will use the solubility product constants (\( K_{sp} \)) for \( \text{AgCl} \) and \( \text{PbCl}_2 \) to find these concentrations.

### Step-by-Step Solution:

1. **Identify the \( K_{sp} \) expressions:**
- For \( \text{AgCl} \):
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
- For \( \text{PbCl}_2 \):
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]

2. **Substitute the known values into the \( K_{sp} \) expressions:**
- For \( \text{AgCl} \):
\[ K_{sp} = 1.8 \times 10^{-10} = [\text{Ag}^+][0.10] \]
Rearranging gives:
\[ [\text{Ag}^+] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \, \text{M} \]

- For \( \text{PbCl}_2 \):
\[ K_{sp} = 1.7 \times 10^{-5} = [\text{Pb}^{2+}][0.10]^2 \]
Rearranging gives:
\[ [\text{Pb}^{2+}] = \frac{1.7 \times 10^{-5}}{0.10 \times 0.10} = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \, \text{M} \]

3. **Final concentrations at equilibrium:**
- The concentration of \( \text{Ag}^+ \) at equilibrium is \( 1.8 \times 10^{-9} \, \text{M} \).
- The concentration of \( \text{Pb}^{2+} \) at equilibrium is \( 1.7 \times 10^{-3} \, \text{M} \).

### Conclusion:
The equilibrium concentrations are:
- \( [\text{Ag}^+] = 1.8 \times 10^{-9} \, \text{M} \)
- \( [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \, \text{M} \)