Solubility of `CdSO_(4)` is `1 xx 10^(-4)` mol per litre. What is the solubility of `CdSO_(4)` in decinormal `H_(2)SO_(4)` solution?

Options

A. `10^(-7)`
B. `2 xx 10^(-7)`
C. `2 xx 10^(-3)`
D. `1 xx 10^(-3)`

Answer

To find the solubility of `CdSO4` in a decinormal `H2SO4` solution, we can follow these steps:

### Step 1: Understand the dissociation of `CdSO4`
When cadmium sulfate (`CdSO4`) dissolves in water, it dissociates into cadmium ions (`Cd^2+`) and sulfate ions (`SO4^2-`):
\[ CdSO4 (s) \rightleftharpoons Cd^{2+} (aq) + SO4^{2-} (aq) \]

### Step 2: Calculate the solubility product constant (Ksp)
Given the solubility of `CdSO4` in pure water is \(1 \times 10^{-4}\) mol/L, we can express the solubility product constant \(K_{sp}\) as follows:
\[ K_{sp} = [Cd^{2+}][SO4^{2-}] = S \times S = (1 \times 10^{-4})^2 = 1 \times 10^{-8} \]

### Step 3: Determine the concentration of sulfate ions in the `H2SO4` solution
In a decinormal (0.1 N) `H2SO4` solution, the normality is related to the molarity by the basicity of sulfuric acid, which is 2 (since it can donate two protons). Therefore, the molarity of `H2SO4` is:
\[ \text{Molarity} = \frac{\text{Normality}}{\text{Basicity}} = \frac{0.1}{2} = 0.05 \, \text{mol/L} \]
Since `H2SO4` dissociates completely, the concentration of sulfate ions (`SO4^2-`) from `H2SO4` is also \(0.05 \, \text{mol/L}\).

### Step 4: Calculate the total concentration of sulfate ions
The total concentration of sulfate ions when `CdSO4` is added to the `H2SO4` solution is the sum of the sulfate ions from both sources:
\[ [SO4^{2-}]_{total} = [SO4^{2-}]_{from \, H2SO4} + [SO4^{2-}]_{from \, CdSO4} = 0.05 + 1 \times 10^{-4} \approx 0.05 \, \text{mol/L} \]
(The contribution from `CdSO4` is negligible compared to the `H2SO4` concentration.)

### Step 5: Set up the expression for solubility in the presence of common ions
The solubility of `CdSO4` in the `H2SO4` solution can be expressed using the \(K_{sp}\) expression:
\[ K_{sp} = [Cd^{2+}][SO4^{2-}]_{total} \]
Let the solubility of `CdSO4` in the `H2SO4` solution be \(S'\). Then:
\[ K_{sp} = S' \times 0.05 \]
Substituting the value of \(K_{sp}\):
\[ 1 \times 10^{-8} = S' \times 0.05 \]

### Step 6: Solve for the solubility \(S'\)
Rearranging the equation to solve for \(S'\):
\[ S' = \frac{1 \times 10^{-8}}{0.05} = 2 \times 10^{-7} \, \text{mol/L} \]

### Final Answer
The solubility of `CdSO4` in a decinormal `H2SO4` solution is:
\[ \boxed{2 \times 10^{-7} \, \text{mol/L}} \]