The solubility product of AgCl at T 298 is 1.8 times 10^-10. For a given solutio
ID: 504657 • Letter: T
Question
The solubility product of AgCl at T 298 is 1.8 times 10^-10. For a given solution with silver concentration [Ag+] = 3 10^-4 what will be the maximum [CI-] concentration before Agcl precipitation? Given that the maximum concentration of Ag_2Cr_4, in water is 6.627 10^-5 M, determine the solubility product of this equilibrium: Ag_2CrO_4 2Ag+ +Cro^2-_4. Calcium Benzoate: A sample of saturated CaBz, solution is at equilibrium: CaBz_2(s) Ca^2+(aq) + 2Bz'(aq). The initially-prepared concentrations are [Ca^2+], = a and [Bz]_i = b. This sample is allowed to saturate, and then the remaining solid is filtered out. As determined by titration, the final benzoate ion concentration in solution is [Bz]_f = c. What is the solubility product of CaBz2, in terms of a, b and c? (Note, this is not simply c2*a!) In our lab experiment, the CaBz_2 solution is saturated at equilibrium before titration. Why must the CaBz_2 solutions be saturated?Explanation / Answer
1)
a)
Ksp of AgCl = 1.8 x 10^-10
AgCl ----------------> Ag+ + Cl-
3.0 x 10^-4 Cl-
Ksp = [Ag+][Cl-]
1.8 x 10^-10 = [3.0 x 10^-4][Cl-]
[Cl-] = 6.0 x 10^-7 M
concentration of [Cl-] = 6.0 x 10^-7 M
b)
conncnetration of Ag2CrO4 = 6.627 x 10^-5
Ag2CrO4 -------------> 2 Ag+ + CrO42-
2S S
Ksp = [Ag+]^2[CrO42-]
= (2S)^2 x S
= 4 S^3
= 4 x (6.627 x 10^-5)^3
Ksp = 1.16 x 10^-12
solubility product = 1.16 x 10^-12