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 HCl to the solution until the concentration of \( \text{Cl}^- \) reaches 0.10 M.

### Step 1: Write the dissociation equations and Ksp expressions

1. For silver chloride (AgCl):
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

2. For lead(II) chloride (PbCl2):
\[ \text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^- \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]

### Step 2: Substitute the known values into the Ksp equations

Given:
- \( K_{sp} \) for \( \text{AgCl} = 1.8 \times 10^{-10} \)
- \( K_{sp} \) for \( \text{PbCl}_2 = 1.7 \times 10^{-5} \)
- \( [\text{Cl}^-] = 0.10 \, \text{M} \)

**For AgCl:**
\[ 1.8 \times 10^{-10} = [\text{Ag}^+][0.10] \]
Rearranging gives:
\[ [\text{Ag}^+] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \, \text{M} \]

**For PbCl2:**
\[ 1.7 \times 10^{-5} = [\text{Pb}^{2+}][0.10]^2 \]
Calculating \( [\text{Cl}^-]^2 \):
\[ 0.10^2 = 0.01 \]
Now substituting:
\[ 1.7 \times 10^{-5} = [\text{Pb}^{2+}](0.01) \]
Rearranging gives:
\[ [\text{Pb}^{2+}] = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \, \text{M} \]

### Step 3: Conclusion

At equilibrium, the concentrations of the ions are:
- \( [\text{Ag}^+] = 1.8 \times 10^{-9} \, \text{M} \)
- \( [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \, \text{M} \)