Question

F. Ksp calculations What is the molar solubility of AgCl given that the Ksp of AgCl is 1.8 x 1010 1. 2. The molar solubility of Ag2CrOs in water is 1.31 x 10. Write the equilibrium expression for the dissociation of silver chromate in water and determine the Ksp for silver chromate.

Answer 1

To find the molar solubility of AgCi Given that Ksp = 1.8x10-10 It will dissociates Aga = Ag* +C1 For every mole of Agathereis one mole of Ag* and one mole of C1" Therefore Ksp=[Ag*[ch] If s= Solubility of Agci Then [Ag*]=s [0]=s Therefore K. 3p = (s) (s) 1.8x10-10 = 52 s2 = 1.8 x 10-20 s=v1.8x10-10 s=1.34x10 PM Hence the molar solubility of Aga = 1.34 x10 M

To find the Ksp for silver chromate (Ag, Croa) Given that molar solubility of Ag, Cro in water is 1.31x104 It will dissociates Ag, Cro. (s) = 2Ag+ (aq) +C+0,2 (aq) For every mole of Ag, CrO, thereistwo moles of Ag* and 1mole of CrO2- Therefore Ko =[Ag*|' [Cr0,2-] K = [2x1.31x104] [1.31x10+] Kg = 8.9x10-12 Hence the K for silver chromate (Ag, Cro ) = 8.9x10-12 |