A saturated 0.01 M `H_(2)` S solution is buffered at pH =3 with exactly sufficient Pb`(NO_(3))_(2)` to not precipitate Pbc. `K_(a) ` of `H_(2) S = 10^(-23),K_(sp)` of Pbs = `10^(-28)` . The concentration of `Pb^(+2)` in the solution is `10^(-x)` . What is x ?

Answer

To solve the problem, we need to find the value of \( x \) such that the concentration of \( \text{Pb}^{2+} \) in the solution is \( 10^{-x} \). We will follow these steps:

### Step 1: Determine the concentration of \( \text{H}^+ \) ions
Given that the pH of the solution is 3, we can calculate the concentration of hydrogen ions (\( \text{H}^+ \)):
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Write the dissociation equation for \( \text{H}_2\text{S} \)
The dissociation of \( \text{H}_2\text{S} \) can be represented as:
\[ \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \]
The dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \]
We know that \( K_a = 10^{-23} \).

### Step 3: Set up the equilibrium expression
Let \( [\text{HS}^-] = x \). At equilibrium, we have:
- \( [\text{H}^+] = 10^{-3} \, \text{M} \)
- \( [\text{H}_2\text{S}] = 0.01 - x \approx 0.01 \) (since \( x \) will be very small compared to 0.01 M)

Substituting these values into the \( K_a \) expression:
\[ 10^{-23} = \frac{(10^{-3})(x)}{0.01} \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ 10^{-23} = \frac{10^{-3}x}{0.01} \implies 10^{-23} = 10^{-1}x \implies x = 10^{-22} \, \text{M} \]
Thus, the concentration of \( \text{HS}^- \) is \( 10^{-22} \, \text{M} \).

### Step 5: Write the dissociation equation for \( \text{PbS} \)
The dissociation of \( \text{PbS} \) can be represented as:
\[ \text{PbS} \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2-} \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{S}^{2-}] \]
Given \( K_{sp} = 10^{-28} \).

### Step 6: Substitute the known values into the \( K_{sp} \) expression
Let \( [\text{Pb}^{2+}] = y \) and we know \( [\text{S}^{2-}] = 10^{-22} \):
\[ 10^{-28} = y \cdot 10^{-22} \]

### Step 7: Solve for \( y \)
Rearranging gives:
\[ y = \frac{10^{-28}}{10^{-22}} = 10^{-6} \, \text{M} \]

### Step 8: Determine \( x \)
Since we have \( [\text{Pb}^{2+}] = 10^{-x} \) and we found \( [\text{Pb}^{2+}] = 10^{-6} \):
\[ 10^{-x} = 10^{-6} \implies x = 6 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{6} \]