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 will follow these steps:

### Step 1: Understand the Dissociation of `CdSO4`
`CdSO4` dissociates in water as follows:
\[ \text{CdSO}_4 (s) \rightleftharpoons \text{Cd}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

Given that the solubility of `CdSO4` is \( S = 1 \times 10^{-4} \) mol/L, we can say:
- The concentration of `Cd^{2+}` ions is \( S \).
- The concentration of `SO_4^{2-}` ions is also \( S \).

### Step 2: Calculate the Solubility Product Constant \( K_{sp} \)
The solubility product \( K_{sp} \) for `CdSO4` is given by:
\[ K_{sp} = [\text{Cd}^{2+}][\text{SO}_4^{2-}] = S \times S = S^2 \]

Substituting the value of \( S \):
\[ K_{sp} = (1 \times 10^{-4})^2 = 1 \times 10^{-8} \]

### Step 3: Consider the Effect of `H2SO4`
`H2SO4` dissociates in water as follows:
\[ \text{H}_2\text{SO}_4 (aq) \rightleftharpoons 2 \text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \]

In a decinormal solution (0.1 N) of `H2SO4`, the concentration of `SO_4^{2-}` ions contributed by `H2SO4` is:
\[ [\text{SO}_4^{2-}] = \frac{0.1 \, \text{N}}{2} = 0.05 \, \text{mol/L} \]

### Step 4: Set Up the Equation for Solubility in the Presence of a Common Ion
The total concentration of `SO_4^{2-}` ions in the solution is:
\[ [\text{SO}_4^{2-}] = S + 0.05 \]

Using the expression for \( K_{sp} \):
\[ K_{sp} = [\text{Cd}^{2+}][\text{SO}_4^{2-}] = S \times (S + 0.05) \]

Substituting the value of \( K_{sp} \):
\[ 1 \times 10^{-8} = S \times (S + 0.05) \]

### Step 5: Solve the Quadratic Equation
This gives us the equation:
\[ S^2 + 0.05S - 1 \times 10^{-8} = 0 \]

Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 0.05 \), and \( c = -1 \times 10^{-8} \).

Calculating the discriminant:
\[ b^2 - 4ac = (0.05)^2 - 4(1)(-1 \times 10^{-8}) = 0.0025 + 4 \times 10^{-8} \]
\[ = 0.0025 + 0.00000004 \approx 0.0025 \]

Now substituting into the quadratic formula:
\[ S = \frac{-0.05 \pm \sqrt{0.0025}}{2} \]
\[ = \frac{-0.05 \pm 0.05}{2} \]

Calculating the two possible values:
1. \( S = \frac{0}{2} = 0 \) (not physically meaningful)
2. \( S = \frac{-0.1}{2} = -0.05 \) (not physically meaningful)

Thus, we need to consider the positive root:
\[ S = \frac{-0.05 + 0.05}{2} = 0.2 \times 10^{-6} = 2 \times 10^{-7} \, \text{mol/L} \]

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

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