At `20^(@)C`, the `Ag^(+)` ion concentration in a saturated solution `Ag_(2)CrO_(4)` is `1.5 x10^(-4)` mol `//` litre. At `20^(@)C`, the solubility product of `Ag_(2)CrO_(4)` would be

Options

A. `1.687xx10^(-12)`
B. `1.75xx10^(-10)`
C. `3.0xx10^(-8)`
D. `4.5xx10^(-10)`

Explanation

To find the solubility product (Ksp) of Ag₂CrO₄ at 20°C, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of silver chromate (Ag₂CrO₄) in water can be represented as:
\[ \text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \]

### Step 2: Identify the concentrations of ions
From the problem, we know that the concentration of Ag⁺ ions in the saturated solution is given as:
\[ [\text{Ag}^+] = 1.5 \times 10^{-4} \text{ mol/L} \]

Since 2 moles of Ag⁺ are produced for every mole of Ag₂CrO₄ that dissolves, the concentration of CrO₄²⁻ ions will be half of the concentration of Ag⁺ ions:
\[ [\text{CrO}_4^{2-}] = \frac{1}{2} [\text{Ag}^+] = \frac{1}{2} \times 1.5 \times 10^{-4} = 0.75 \times 10^{-4} \text{ mol/L} \]

### Step 3: Write the expression for Ksp
The solubility product (Ksp) expression for Ag₂CrO₄ is given by:
\[ K_{sp} = [\text{Ag}^+]^2 \times [\text{CrO}_4^{2-}] \]

### Step 4: Substitute the concentrations into the Ksp expression
Now we can substitute the values we have:
\[ K_{sp} = (1.5 \times 10^{-4})^2 \times (0.75 \times 10^{-4}) \]

### Step 5: Calculate Ksp
First, calculate \((1.5 \times 10^{-4})^2\):
\[ (1.5 \times 10^{-4})^2 = 2.25 \times 10^{-8} \]

Now, multiply this by \((0.75 \times 10^{-4})\):
\[ K_{sp} = 2.25 \times 10^{-8} \times 0.75 \times 10^{-4} \]
\[ K_{sp} = 1.6875 \times 10^{-12} \]

### Conclusion
Thus, the solubility product \( K_{sp} \) of Ag₂CrO₄ at 20°C is:
\[ K_{sp} = 1.6875 \times 10^{-12} \]

### Final Answer
The answer is approximately:
\[ K_{sp} \approx 1.688 \times 10^{-12} \]

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