Solubility of `Ag_(2)CrO_(4)` (`K_(sp)=4xx10^(-13)`) in 0.1 M `K_(2)CrO_(4)` solution will be :

Options

A. `10^(-3)` M
B. `10^(-6)` M
C. `4xx10^(6)` M
D. `5xx10^(-7)` M

Answer

To find the solubility of `Ag2CrO4` in a `0.1 M K2CrO4` solution, we can follow these steps:

### Step 1: Write the dissociation equation for `Ag2CrO4`
The dissociation of silver chromate in water can be represented as:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2 Ag^+ (aq) + CrO_4^{2-} (aq) \]

### Step 2: Write the expression for the solubility product constant (`Ksp`)
The solubility product constant (`Ksp`) for this reaction is given by:
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \]
Given that \( K_{sp} = 4 \times 10^{-13} \).

### Step 3: Determine the concentration of `CrO4^{2-}` in the solution
Since we are dissolving `Ag2CrO4` in a `0.1 M K2CrO4` solution, the concentration of `CrO4^{2-}` ions is already provided by the potassium chromate:
\[ [CrO_4^{2-}] = 0.1 \, M \]

### Step 4: Set up the equation for the solubility of `Ag2CrO4`
Let the solubility of `Ag2CrO4` in this solution be \( s \). From the dissociation equation, for every mole of `Ag2CrO4` that dissolves, 2 moles of `Ag^+` are produced. Therefore:
\[ [Ag^+] = 2s \]

### Step 5: Substitute the concentrations into the `Ksp` expression
Now substitute the concentrations into the `Ksp` expression:
\[ K_{sp} = (2s)^2 \cdot (0.1) \]
\[ 4 \times 10^{-13} = 4s^2 \cdot 0.1 \]

### Step 6: Solve for `s`
Rearranging the equation gives:
\[ 4 \times 10^{-13} = 0.4s^2 \]
\[ s^2 = \frac{4 \times 10^{-13}}{0.4} \]
\[ s^2 = 1 \times 10^{-12} \]
Taking the square root:
\[ s = 1 \times 10^{-6} \, M \]

### Conclusion
The solubility of `Ag2CrO4` in a `0.1 M K2CrO4` solution is:
\[ s = 1 \times 10^{-6} \, M \]

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