Calculate the molar solubility of silver chromate in pure water. Ksp = 1.12 x 10¯12
Solubility equilibrium of Ag2CrO4(s) is
Ag2CrO4(s) <--------> 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-] = 1.12 ×10-12
If solubility of Ag2CrO4 is represented by S ,
at saturated solution
[Ag+] = 2S
[CrO42-] = S
(2S)2S = 1.12 ×10-12
4S3 = 1.12 ×10-12
S3 = 2.8 ×10-13
S = 6.54 ×10-5
Therefore
Molar solubility of Ag2CrO4 = 6.54 ×10-5 mol/L
Calculate the molar solubility of silver chromate in pure water. Ksp = 1.12 x 10¯12
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