Calculating the Effect of a Common Ion on Solubility
Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-]
Concentration (M) Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq)
Initial --------- 0.0600 0
Change --------- +2x +x
Equilibrium --------- 0.0600 + 2x x
Assuming that Ksp is small, 0.0600 M + 2x = 0.600 M
Ksp = 2.6 x 10-12 = (0.0600)2(x) x = 7.22 x 10-10 M
Therefore, the solubility of silver chromate is 7.22 x 10-10 M
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