Find the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of the cell Ag/Ag+ (saturated Ag2CrO4) || Ag+ (0.1 M)/Ag is 0.164 V at 298.

To solve this problem, we need to use the Nernst equation and the solubility product expression.

Step 1: Write down the Nernst equation The Nernst equation is E = E° - (RT/nF)lnQ, where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient.

Step 2: Identify the values in the problem From the problem, we know that E = 0.164 V, T = 298 K, and the concentration of Ag+ in the solution is 0.1 M. We also know that the reaction is Ag2CrO4 (s) ⇌ 2Ag+ (aq) + CrO4 2- (aq), so n = 2.

Step 3: Calculate E° We know that E° for the Ag+/Ag half-cell is 0.8 V. Since the cell potential is the difference in potentials of the two half-cells and the other half-cell is a saturated solution of Ag2CrO4, E° for the cell is 0.8 V.

Step 4: Substitute the values into the Nernst equation and solve for Q 0.164 V = 0.8 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol) * lnQ Solving for Q gives Q = [Ag+]^2 / [CrO4 2-] = 10^-10

Step 5: Write the solubility product expression and solve for Ksp The solubility product expression is Ksp = [Ag+]^2 * [CrO4 2-]. Since the solution is saturated, [Ag+] = [CrO4 2-], so Ksp = [Ag+]^3. Substituting Q into the expression gives Ksp = (10^-10)^3/2 = 10^-15.

So, the solubility product of Ag2CrO4 at 298 K is 10^-15.