Question

The solubility of AgCl in water is 1.34x10-M, but increases in the presence of thiosulfate ion. Calculate the solubility of AgCl when the solid is added to a solution that is 0.704 M in thiosulfate ion. For AgCI, Ksp = 1.80x10-10 and for Ag(S203)23, Kr=2.00x1013 Solubility =

Answer 1

Complete reactions are

AgCl (s)----> Ag⁺(aq) + Cl^-(aq) .... K_sp = 1.80 *10^-10 M

Ag⁺ + 2S₂O₃^2- ----> [Ag(S₂O₃)₂]^3- .... K_f = 2.00 *10¹³

adding above two, we get

AgCl + 2S₂O₃^2- ----> [Ag(S₂O₃)₂]^3- + Cl^- .. Keq = K_s *K_f = 1.80 *10^-10 * 2.00 *10¹³ = 3600

	AgCl	2S2O32-	---->	[Ag(S2O3)2]3-	Cl-
I(M)		0.704		0	1.34 *10-5
C(M)		-2x		x	+x
E(M)		0.704-2x		x	(1.34 *10-5)+x

K_f = [Ag(S₂O₃)₂]^3- [Cl^-] / [S₂O₃^2-]²

K_f = (x* ((1.34 *10^-5)+x)) / (0.704- 2x)²

3600 = (x² + (1.34 *10^-5x) / (0.496 + 4x² - 2.816x)

x =0.349 M

Thus, Solubility = 0.349 M