If for a binary weak electrolyte the solubility product is `4xx10^(-10) ` at 298K.Calculate its solubility is mol `dm^(-3)` at the same temperature

Options

A. `4xx10^(-5)`
B. `2xx10^(-5)`
C. `8xx10^(-10)`
D. `16xx10^(-20)`

Answer

To solve the problem of finding the solubility of a binary weak electrolyte given its solubility product (Ksp), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Electrolyte:**
- Let's denote the binary weak electrolyte as AB. When it dissolves in water, it dissociates into its cation (A⁺) and anion (B⁻).
- The dissociation can be represented as:
\[ AB \rightleftharpoons A^+ + B^- \]

2. **Setting Up the Solubility Expression:**
- Let the solubility of the electrolyte AB be \( S \) mol/dm³.
- At equilibrium, the concentration of A⁺ and B⁻ will both be equal to \( S \) mol/dm³.

3. **Writing the Solubility Product Expression:**
- The solubility product \( K_{sp} \) for the dissociation is given by:
\[ K_{sp} = [A^+][B^-] \]
- Since both ions have the same concentration at equilibrium, we can write:
\[ K_{sp} = S \times S = S^2 \]

4. **Substituting the Given Ksp Value:**
- We know that \( K_{sp} = 4 \times 10^{-10} \).
- Therefore, we can set up the equation:
\[ S^2 = 4 \times 10^{-10} \]

5. **Calculating the Solubility (S):**
- To find \( S \), we take the square root of both sides:
\[ S = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \text{ mol/dm}^3 \]

6. **Conclusion:**
- The solubility of the binary weak electrolyte at 298 K is:
\[ S = 2 \times 10^{-5} \text{ mol/dm}^3 \]

### Final Answer:
The solubility of the binary weak electrolyte is \( 2 \times 10^{-5} \) mol/dm³.

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