In: Chemistry
Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 3.573 using the systematic treatment of equilibrium. Ksp(Ag2CO3) = 8.46 × 10–12; Ka1(H2CO3) = 4.45 × 10–7; Ka2(H2CO3) = 4.69 × 10–11. Please show step as to how you got the answer, i would really appreciate it.
Ag2CO3(s) <==> 2Ag+ + CO3^2- ------ Ksp
CO3^2- + H+ <===> HCO3- ...............1/Ka2
adding borth equations,
Ag2CO3(s) + H+ <===> 2Ag+ + HCO3-
K = 8.46 x 10^-12/4.69 x 10^-11 = 0.180
pH = -log[H+] = 3.573
[H+] = 2.7 x 10^-4 M
ICE chart
Ag2CO3(s) + H+ <===> 2Ag+ + HCO3-
I - 2.7 x 10^-4 - -
C - -x +x +2x
E - 2.7 x 10^-4 - x x 2x
So,
K = [HCO3-][Ag+]^2/[H+]
0.180 = (x)(2x)^2/(2.7 x 10^-4 - x)
let x be a small amount then,
0.180 = (x)(2x)^2/(2.7 x 10^-4)
4x^3 = 4.86 x 10^-5
x = 0.03 M
so molar solubility of Ag2CO3 was found to be 0.03 M