1.sol:
Ag2CrO4 = 2Ag+ + CrO42-
s 2s s
Ksp = [2s]2[s] = 4s3 , Ksp= 1.1 x 10-12 (given)
s3 = Ksp/4 = 1.1 x 10-12 / 4 = 0.275 x 10-2
s3= 0.275 x 10-2
3logs = log(0.275 x 10-2) ; log s = -0.8536 ; s = antilog(-0.8536) = 14.01 x 10-2 mol/L
hence , s = 14.01 x 10-2 mol/L
S = 14.01 x 10-2 x 331.74g/L = 46.48g/L , (Ag2CrO4 mol.wt. = 331.74 g/mol)
hence, 46.48g Ag2CrO4 will dissolve in 1L water.
2.ans Data value for H , S ,G calculation not given.
F the solubility product 1. How many grams of Ag CrO4 will dissolve in pure water i constant for ...
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