Question

Calculate the [Ag⁺] in a solution prepared by dissolving 1.00g of AgNO₃ and 10.0g KCN in enough water to make a 1.00 L of solution Kf[Ag(CN)₂]^1-=1.0 x 10²¹

Answer: [Ag⁺] = 2.9 x 10^-22M

Answer 1

Given,

Mass of AgNO₃ = 1.00 g

Mass of KCN = 10.0 g

Volume of solution = 1.00 L

K_f value for [Ag(CN)₂]^- = 1.0 x 10²¹

Calculating the number of moles of AgNO₃ and KCn from the given masses,

= 1.00 g of AgNO₃ x (1 mol /169.87 g)

= 0.005887 mol of AgNO₃

= 0.005887 mol of Ag⁺

Similarly,

= 10.0 g of KCN x ( 1 mol / 65.12 g)

= 0.1536 mol of KCN

= 0.1536 mol of CN^-

Now, the reaction between Ag⁺ and CN^- is,

Ag⁺(aq) + 2CN^-(aq) [Ag(CN)₂]^-(aq)

Drawing an ICE chart,

	Ag+(aq)	2CN-(aq)	[Ag(CN)2]-(aq)
I(moles)	0.005887	0.1536	0
C(moles)	-0.005887	-2(0.005887)	+0.005887
E(moles)	0	0.1418	0.005887

Now, the new concentration for [CN^-] and [Ag(CN)₂]^-

[CN^-] = 0.1418 mol /1 L = 0.1418 M

[Ag(CN)₂]^- = 0.005887 mol / 1 L = 0.005887 M

Now, the dissociation reaction of [Ag(CN)₂]^-,

[Ag(CN)₂]^-(aq) Ag⁺(aq) + 2CN^-(aq)

	[Ag(CN)2]-(aq)	Ag+(aq)	2CN-(aq)
I(M)	0.005887	0	0.1418
C(M)	-x	+x	+2x
E(M)	0.005887-x	x	0.1418+2x

Now, K_d expression,

K_d = [CN^-]² [Ag⁺] /[[Ag(CN)₂]^-]

K_d = 1 /K_f

K_d = 1 /(1.0x10²¹)

K_d = 1.0 x 10^-21

1.0 x 10^-21 = [0.1418 +2x]² [x] /[0.005887-x]

1.0 x 10^-21 = [0.1418] [x] /[0.005887] -------[0.1418 +2x] 0.1418 M Or [0.005887-x] 0.005887, x<<0.14,0.0059

x = 2.9 x 10^-22

Thus, [Ag⁺] = x = 2.9 x 10^-22 M