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. `3.3750 xx 10^(-12)`
B. `1.6875 xx 10^(-10)`
C. `1.6875 xx 10^(-12)`
D. `1.6875 xx 10^(-11)`

Explanation

To find the solubility product (Ksp) of silver chromate (Ag₂CrO₄), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of silver chromate 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: Determine the concentrations of ions
From the dissociation equation, we can see that for every 1 mole of Ag₂CrO₄ that dissolves, it produces 2 moles of Ag⁺ ions and 1 mole of CrO₄²⁻ ions.

Given that the concentration of Ag⁺ ions in the saturated solution is \(1.5 \times 10^{-4}\) mol/L, we can find the concentration of CrO₄²⁻ ions.

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

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

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

### Step 5: Calculate Ksp
Calculating the values:
1. Calculate \((1.5 \times 10^{-4})^2\):
\[ (1.5 \times 10^{-4})^2 = 2.25 \times 10^{-8} \]
2. Now multiply by \((0.75 \times 10^{-4})\):
\[ K_{sp} = 2.25 \times 10^{-8} \times 0.75 \times 10^{-4} = 1.6875 \times 10^{-11} \]

### Final Answer
Thus, the solubility product \(K_{sp}\) of Ag₂CrO₄ at \(20^{\circ}C\) is:
\[ K_{sp} = 1.6875 \times 10^{-11} \]

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