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 S2O32-. For [Ag(S2O3)2]3-, Kf = 2.9×1013.
[Ag+] = ____M
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
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