If s is the molar solubility of `Ag_2SO_4`, then

Options

A. `3[Ag^+]=s`
B. `[Ag^+]=s`
C. `2[Ag^+]=s`
D. `[SO_(4)^(2-)]=s`

Answer

To solve the problem regarding the molar solubility of silver sulfate (Ag₂SO₄), we need to analyze the dissociation of the compound in water and relate it to its solubility product (Ksp).

### Step-by-Step Solution:

1. **Dissociation of Ag₂SO₄**:
When silver sulfate (Ag₂SO₄) dissolves in water, it dissociates into its constituent ions:
\[ \text{Ag}_2\text{SO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{SO}_4^{2-} (aq) \]
This means that for every mole of Ag₂SO₄ that dissolves, it produces 2 moles of silver ions (Ag⁺) and 1 mole of sulfate ions (SO₄²⁻).

2. **Define Molar Solubility (S)**:
Let \( S \) be the molar solubility of Ag₂SO₄. This means that if \( S \) moles of Ag₂SO₄ dissolve, then:
- The concentration of Ag⁺ ions will be \( 2S \) (since 2 moles of Ag⁺ are produced for every mole of Ag₂SO₄).
- The concentration of SO₄²⁻ ions will be \( S \).

3. **Expression for Ksp**:
The solubility product constant (Ksp) for the dissociation of Ag₂SO₄ can be expressed as:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{SO}_4^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 \cdot (S) = 4S^2 \cdot S = 4S^3 \]

4. **Conclusion**:
From the above analysis, we can conclude that the relationship between the molar solubility \( S \) and the concentrations of the ions in solution is:
- Concentration of Ag⁺ = \( 2S \)
- Concentration of SO₄²⁻ = \( S \)

### Final Answer:
Thus, the correct relationship is that the concentration of sulfate ions (SO₄²⁻) is equal to \( S \), while the concentration of silver ions (Ag⁺) is \( 2S \).