Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
`AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)]`
Where `K_(sp)` is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as `K_(sp)` are
`QltK_(sp)` Unsaturated solution
`Q=K_(sp)` Saturated solution
`Qgt_(sp)` Supersaturated solution, precipitate will from
Will a precipitate from if 1 volume of 0.1 volume of 0.1 `MPb^(2+)` ion solution in mixed with 3 volume of 0.3 M `Cl^(-)` ion solution ? `["Givem":K_(sp)(PbCl_(2))=1.7xx10^(-5)M^(3)]`

Options

A. Yes
B. No
C. Ionic product is less than solubility product, hence precipitate will from
D. Data insufficient

Answer

To determine whether a precipitate will form when mixing the given solutions, we can follow these steps:

### Step 1: Determine the moles of Pb²⁺ and Cl⁻ ions in the mixed solution.
- **Volume of Pb²⁺ solution:** 0.1 V (where V is the volume unit)
- **Concentration of Pb²⁺ solution:** 0.1 M
- **Moles of Pb²⁺:**
\[ \text{Moles of Pb}^{2+} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.1 V = 0.01 V \, \text{moles} \]

- **Volume of Cl⁻ solution:** 3 V
- **Concentration of Cl⁻ solution:** 0.3 M
- **Moles of Cl⁻:**
\[ \text{Moles of Cl}^- = \text{Concentration} \times \text{Volume} = 0.3 \, \text{M} \times 3 V = 0.9 V \, \text{moles} \]

### Step 2: Calculate the total volume of the mixed solution.
- **Total volume (V_total):**
\[ V_{\text{total}} = 0.1 V + 3 V = 3.1 V \]

### Step 3: Calculate the concentrations of Pb²⁺ and Cl⁻ in the mixed solution.
- **Concentration of Pb²⁺:**
\[ [\text{Pb}^{2+}] = \frac{\text{Moles of Pb}^{2+}}{V_{\text{total}}} = \frac{0.01 V}{3.1 V} = \frac{0.01}{3.1} \, \text{M} \approx 3.23 \times 10^{-3} \, \text{M} \]

- **Concentration of Cl⁻:**
\[ [\text{Cl}^-] = \frac{\text{Moles of Cl}^-}{V_{\text{total}}} = \frac{0.9 V}{3.1 V} = \frac{0.9}{3.1} \, \text{M} \approx 0.2903 \, \text{M} \]

### Step 4: Calculate the reaction quotient (Q) for the precipitation of PbCl₂.
- The equilibrium expression for the dissolution of PbCl₂ is:
\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2 \text{Cl}^-(aq) \]
- The reaction quotient (Q) is given by:
\[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
- Substituting the concentrations:
\[ Q = (3.23 \times 10^{-3}) \times (0.2903)^2 \]
\[ Q \approx (3.23 \times 10^{-3}) \times (0.0843) \approx 2.73 \times 10^{-4} \]

### Step 5: Compare Q with K_sp to determine if a precipitate will form.
- Given \( K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5} \, \text{M}^3 \)
- Since \( Q = 2.73 \times 10^{-4} \) and \( K_{sp} = 1.7 \times 10^{-5} \):
\[ Q > K_{sp} \]
- Therefore, a precipitate of PbCl₂ will form.

### Conclusion:
Yes, a precipitate will form when the two solutions are mixed.

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