The `K_(sp) ` of `PbCO_(3) and MgCO_(3)` are `1.5xx10^(-15) and 1xx10^(-15)` respectively at 298 K. The concentration of `Pb^(2+)` ions in a saturated solution containing `MgCO_(3) and PbCO_(3)` is

Options

A. `1.5 xx 10^(-8)M`
B. `3xx10^(-8) M`
C. `2xx10^(-8) M`
D. `2.5xx10^(-8)M`

Explanation

To find the concentration of \( \text{Pb}^{2+} \) ions in a saturated solution containing both \( \text{MgCO}_3 \) and \( \text{PbCO}_3 \), we can follow these steps:

### Step 1: Write the dissociation equations
For \( \text{PbCO}_3 \):
\[ \text{PbCO}_3 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]
For \( \text{MgCO}_3 \):
\[ \text{MgCO}_3 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]

### Step 2: Define solubility
Let the solubility of \( \text{PbCO}_3 \) be \( x \) and that of \( \text{MgCO}_3 \) be \( y \).

From the dissociation equations:
- The concentration of \( \text{Pb}^{2+} \) will be \( x \).
- The concentration of \( \text{CO}_3^{2-} \) will be \( x + y \) (since both salts contribute to the carbonate ion concentration).

### Step 3: Write the \( K_{sp} \) expressions
The solubility product constant \( K_{sp} \) for \( \text{PbCO}_3 \) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{CO}_3^{2-}] = x(x + y) \]
Given \( K_{sp} \) for \( \text{PbCO}_3 = 1.5 \times 10^{-15} \), we have:
\[ 1.5 \times 10^{-15} = x(x + y) \quad \text{(1)} \]

For \( \text{MgCO}_3 \):
\[ K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}] = y(x + y) \]
Given \( K_{sp} \) for \( \text{MgCO}_3 = 1.0 \times 10^{-15} \), we have:
\[ 1.0 \times 10^{-15} = y(x + y) \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ 1.5 \times 10^{-15} = x(x + y) \]

From equation (2):
\[ 1.0 \times 10^{-15} = y(x + y) \]

### Step 5: Divide the two equations
Dividing equation (1) by equation (2):
\[ \frac{1.5 \times 10^{-15}}{1.0 \times 10^{-15}} = \frac{x(x + y)}{y(x + y)} \]
This simplifies to:
\[ 1.5 = \frac{x}{y} \]
Thus, we can express \( x \) in terms of \( y \):
\[ x = 1.5y \quad \text{(3)} \]

### Step 6: Substitute \( x \) into one of the equations
Substituting equation (3) into equation (1):
\[ 1.5 \times 10^{-15} = (1.5y)(1.5y + y) \]
This simplifies to:
\[ 1.5 \times 10^{-15} = (1.5y)(2.5y) = 3.75y^2 \]

### Step 7: Solve for \( y \)
Rearranging gives:
\[ y^2 = \frac{1.5 \times 10^{-15}}{3.75} \]
Calculating this:
\[ y^2 = 4 \times 10^{-16} \quad \Rightarrow \quad y = 2 \times 10^{-8} \]

### Step 8: Find \( x \)
Using equation (3):
\[ x = 1.5y = 1.5 \times (2 \times 10^{-8}) = 3 \times 10^{-8} \]

### Conclusion
The concentration of \( \text{Pb}^{2+} \) ions in the saturated solution is:
\[ \boxed{3 \times 10^{-8} \, \text{M}} \]