The solubility product of `Pbl_(2)` is `7.2 xx 10^(-9)`. The maximum mass of `Nal` which may be added in `500ml` of `0.005M Pb (NO_(3))_(2)` solution without any precipitation of `Pbl_(2)` is `( = 127)`:

Options

A. `0.09g`
B. `1.2 xx 10^(-3)g`
C. `6 xx 10^(-4)g`
D. `1.08 xx 10^(-5)g`

Explanation

To solve the problem, we need to determine the maximum mass of NaI that can be added to a 500 ml solution of 0.005 M Pb(NO3)2 without causing precipitation of PbI2. The solubility product (Ksp) of PbI2 is given as 7.2 x 10^(-9).

### Step-by-Step Solution:

1. **Write the Dissociation Reaction of PbI2:**
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

2. **Write the Expression for the Solubility Product (Ksp):**
The expression for Ksp is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]

3. **Substitute Known Values:**
We know:
- \( K_{sp} = 7.2 \times 10^{-9} \)
- The concentration of \( \text{Pb}^{2+} \) from the Pb(NO3)2 solution is 0.005 M.

Substituting these values into the Ksp expression:
\[ 7.2 \times 10^{-9} = (0.005)[\text{I}^-]^2 \]

4. **Solve for \([\text{I}^-]\):**
Rearranging the equation to find \([\text{I}^-]^2\):
\[ [\text{I}^-]^2 = \frac{7.2 \times 10^{-9}}{0.005} \]
\[ [\text{I}^-]^2 = 1.44 \times 10^{-6} \]
Taking the square root:
\[ [\text{I}^-] = \sqrt{1.44 \times 10^{-6}} = 1.2 \times 10^{-3} \text{ M} \]

5. **Calculate the Number of Moles of I- in 500 ml:**
The volume of the solution is 500 ml, which is 0.5 L. The number of moles of \([\text{I}^-]\) can be calculated as:
\[ \text{Moles of I}^- = [\text{I}^-] \times \text{Volume} = (1.2 \times 10^{-3} \text{ M}) \times (0.5 \text{ L}) = 6 \times 10^{-4} \text{ moles} \]

6. **Convert Moles of I- to Mass of NaI:**
The molar mass of NaI is approximately 150 g/mol. Therefore, the mass of NaI can be calculated as:
\[ \text{Mass of NaI} = \text{Moles of I}^- \times \text{Molar Mass of NaI} = (6 \times 10^{-4} \text{ moles}) \times (150 \text{ g/mol}) = 0.09 \text{ g} \]

### Final Answer:
The maximum mass of NaI that can be added without causing precipitation of PbI2 is **0.09 g**.