Question

What is the approximate concentration of free Ag⁺ ion at equilibrium when 1.42×10^-2 mol silver nitrate is added to 1.00 L of solution that is 1.070 M in S₂O₃^2-. For [Ag(S₂O₃)₂]^3-, K_f = 2.9×10¹³.

[Ag⁺] = ____M

Answer 1

the concentrations:

[AgNO3] = [Ag+] = 1.42*10^-2 = 0.0142 M

[S2O3-2] = 1.070 M

Recall that

Ag+ + 2S2O3-2 <--> Ag(S2O3)2-3

Kf = [Ag(S2O3)2-3] / [Ag+][S2O3-]^2

initially

[Ag+] = 0.0142

[S2O3-2] = 1.070

[Ag(S2O3)2-3] = 0

in equilbirium

[Ag+] = 0.0142 - x

[S2O3-2] = 1.070 -2x

[Ag(S2O3)2-3] = 0 + x

substitute in Kf

Kf = [Ag(S2O3)2-3] / [Ag+][S2O3-]^2

2.9*10^13 = (x) / ((0.0142 - x)(1.070 -2x))

2.9*10^13 = (x) / ( 0.0142*1.070 - (0.0142*2 + 1.070 )x + 2x^2)

( 0.0142*1.070 - (0.0142*2 + 1.070 )x + 2x^2) = x/2.9*10^13

( 0.0142*1.070 - (0.0142*2 + 1.070 )x + 2x^2) = 0

2x^2 -1.0984x + 0.015194 = 0

x = 0.01419

[Ag+] = 0.0142 - x = 0.142-0.01419 = 0.12781 M

[S2O3-2] = 1.070 -2x = 1.070-2*0.01419 = 1.04162 M

[Ag(S2O3)2-3] = 0 + x = 0.01419

free [Ag+] =  0.12781 M