What is the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the EMF of the cell: Ag \| Ag^+ (satd. Ag2CrO4) \|\| Ag^+ (0.1 M) \| Ag is 0.162 V at 298 K? [2.303RT/F = 0.06, l

Question

What is the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the EMF of the cell: Ag \| Ag^+ (satd. Ag2CrO4) \|\| Ag^+ (0.1 M) \| Ag is 0.162 V at 298 K? [2.303RT/F = 0.06, log 2 = 0.3]

Prashant Kotian · Accepted Answer

Correct answer: D. E_cell = 0.06 * log(0.1 / [Ag^+]_anode). 0.162 = 0.06 * log(0.1 / [Ag^+]_a), so log(0.1 / [Ag^+]_a) = 2.7. 0.1 / [Ag^+]_a = 10^2.7 = 500. [Ag^+]_a = 0.1 / 500 = 2 * 10^-4 M. For Ag2CrO4, [CrO4^2-] = 1 * 10^-4 M. Ksp = [Ag^+]^2 [CrO4^2-] = (2 * 10^-4)^2 * (1 * 10^-4) = 4.0 * 10^-12. Therefore, correct answer is D.

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