`PbCl_(2)` has a solubility product of `1.7xx10^(-8)`. Will a precipitate of `PbCi_(2)` form when 0.010 mole of lead nitrate and 0.010 mole of potassium chloride are mixed and water added upto 1 litre ?

Explanation

To determine whether a precipitate of lead(II) chloride (PbCl₂) will form when mixing lead nitrate (Pb(NO₃)₂) and potassium chloride (KCl), we can follow these steps:

### Step 1: Write down the solubility product (Ksp) of PbCl₂
The solubility product (Ksp) of PbCl₂ is given as:
\[ K_{sp} = 1.7 \times 10^{-8} \]

### Step 2: Determine the initial concentrations of Pb²⁺ and Cl⁻ ions
When 0.010 moles of lead nitrate (Pb(NO₃)₂) and 0.010 moles of potassium chloride (KCl) are mixed and diluted to a total volume of 1 liter, we can find the concentrations of the ions.

- Lead nitrate dissociates into Pb²⁺ and 2 NO₃⁻:
\[ \text{Pb(NO}_3\text{)}_2 \rightarrow \text{Pb}^{2+} + 2 \text{NO}_3^{-} \]
Therefore, the concentration of Pb²⁺ is:
\[ [\text{Pb}^{2+}] = \frac{0.010 \text{ moles}}{1 \text{ L}} = 0.010 \, \text{M} \]

- Potassium chloride dissociates into K⁺ and Cl⁻:
\[ \text{KCl} \rightarrow \text{K}^{+} + \text{Cl}^{-} \]
Therefore, the concentration of Cl⁻ is:
\[ [\text{Cl}^{-}] = \frac{0.010 \text{ moles}}{1 \text{ L}} = 0.010 \, \text{M} \]

### Step 3: Calculate the ionic product (IP) for PbCl₂
The ionic product (IP) for the precipitation of PbCl₂ is calculated using the formula:
\[ IP = [\text{Pb}^{2+}] \times [\text{Cl}^{-}]^2 \]
Substituting the values:
\[ IP = (0.010) \times (0.010)^2 = 0.010 \times 0.0001 = 1.0 \times 10^{-6} \]

### Step 4: Compare the ionic product (IP) with the solubility product (Ksp)
Now we compare the calculated ionic product with the given Ksp:
- \( IP = 1.0 \times 10^{-6} \)
- \( K_{sp} = 1.7 \times 10^{-8} \)

Since \( IP > K_{sp} \), this indicates that the solution is supersaturated with respect to PbCl₂.

### Step 5: Conclusion
Since the ionic product exceeds the solubility product, a precipitate of PbCl₂ will form.

### Final Answer
Yes, a precipitate of PbCl₂ will form.

---