Find the solubility product of a saturated solution of `Ag_(2)CrO_(4)` in water at `298K`, if the `EMF` of the cell `:`
`Ag|Ag^(o+)(satAg_(2)CrO_(4)sol)||Ag(0.1M)|Agis 0.164V` at `298K`.

Explanation

To find the solubility product (Ksp) of a saturated solution of \( \text{Ag}_2\text{CrO}_4 \) in water at 298 K, we can follow these steps:

### Step 1: Understand the Cell and EMF
We have a galvanic cell represented as:
\[ \text{Ag} | \text{Ag}^+ (\text{sat } \text{Ag}_2\text{CrO}_4) || \text{Ag}^+ (0.1 \, \text{M}) | \text{Ag} \]
The EMF of the cell is given as \( E = 0.164 \, \text{V} \).

### Step 2: Use the Nernst Equation
The Nernst equation relates the cell potential to the concentrations of the ions involved:
\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{C_2}{C_1} \right) \]
Where:
- \( E \) is the cell potential (0.164 V)
- \( E^0 \) is the standard cell potential (to be determined)
- \( n \) is the number of electrons transferred (for \( \text{Ag}^+ \) to \( \text{Ag} \), \( n = 1 \))
- \( C_1 \) is the concentration of \( \text{Ag}^+ \) in the saturated solution
- \( C_2 \) is the concentration of \( \text{Ag}^+ \) in the 0.1 M solution

### Step 3: Rearranging the Nernst Equation
Rearranging the equation gives:
\[ E^0 = E + \frac{0.0591}{n} \log \left( \frac{C_2}{C_1} \right) \]

### Step 4: Substitute Known Values
Substituting \( E = 0.164 \, \text{V} \), \( C_2 = 0.1 \, \text{M} \), and \( n = 1 \):
\[ E^0 = 0.164 + \frac{0.0591}{1} \log \left( \frac{0.1}{C_1} \right) \]

### Step 5: Solve for \( C_1 \)
We can express \( C_1 \) in terms of \( E^0 \):
1. Calculate \( \frac{0.164}{0.0591} \):
\[ \frac{0.164}{0.0591} \approx 2.774 \]
2. Set up the equation:
\[ 2.774 = \log \left( \frac{0.1}{C_1} \right) \]
3. Convert from logarithmic form:
\[ \frac{0.1}{C_1} = 10^{2.774} \]
\[ C_1 = \frac{0.1}{10^{2.774}} \approx 1.66 \times 10^{-4} \, \text{M} \]

### Step 6: Determine Ion Concentrations
From the dissociation of \( \text{Ag}_2\text{CrO}_4 \):
\[ \text{Ag}_2\text{CrO}_4 \rightleftharpoons 2 \text{Ag}^+ + \text{CrO}_4^{2-} \]
Let \( s \) be the solubility of \( \text{Ag}_2\text{CrO}_4 \):
- \( [\text{Ag}^+] = 2s \)
- \( [\text{CrO}_4^{2-}] = s \)

From \( C_1 \):
\[ 2s = 1.66 \times 10^{-4} \implies s = \frac{1.66 \times 10^{-4}}{2} = 0.83 \times 10^{-4} \, \text{M} \]

### Step 7: Calculate Ksp
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
Substituting the values:
\[ K_{sp} = (1.66 \times 10^{-4})^2 \times (0.83 \times 10^{-4}) \]
Calculating:
\[ K_{sp} \approx 2.287 \times 10^{-12} \]

### Final Answer
The solubility product of \( \text{Ag}_2\text{CrO}_4 \) at 298 K is approximately:
\[ \boxed{2.287 \times 10^{-12}} \]