Solubility equilibrium of Ag2CrO4 is
Ag2CrO4(s) <-------> 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-] = 1.10 ×10-12
substitute the given concentration of CrO42- ion in Ksp expression
[Ag+]2 × 0.0346M = 1.10 ×10-12 M3
[Ag+]2 = 3.179 × 10-11M2
[Ag+] = 5.64×10-6M
Therefore,
Concentration of silver ion required to just initiate precipitate = 5.64 ×10-6M
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