Question

Determine the molar concentration of Ag+(aq) ions in a 1.00M solution of AgCl2^- with no excess chloride

Given: Kf= 2.50x10^5 for AgCl2^-

Answer 1

Given,

Concentration of [AgCl₂]^- = 1.00 M

K_f of [AgCl₂]^- = 2.50 x 10⁵

Now, the dissociation reaction for [AgCl₂]^- is,

[AgCl₂]^- (aq) Ag⁺(aq) + 2Cl^-(aq)

Drawing an ICE chart,

	[AgCl2]- (aq)	Ag+(aq)	2Cl-(aq)
I(M)	1.00	0	0
C(M)	-x	+x	+2x
E(M)	1.00-x	x	2x

Now, the K_d expression is,

K_d = [Cl^-]² [Ag⁺] / [[AgCl₂]^-]

Now, converting K_f to K_d,

K_d = 1 / K_f [ Since, we reversed the formation reaction]

K_d = 1 / 2.50 x 10⁵

K_d = 4.0 x 10^-6

Now,

4.0 x 10^-6 = [2x]² [x] / [1.00-x]

4.0 x 10^-6 = 4x³ / [1.00] ------------Here, [1.00-x] 1.00 since, x<<<1.00

x = 0.01

Thus, [Ag⁺] = x = 0.01 M