For the reaction, #["Ag(CN)"_2]^(-)-> "Ag"^+ + "2CN"^-# the equilibrium constant at 298 K is #4.0 × 10^-10#. The silver ion concentration in a solution which was originally 0.12 M in KCN and 0.04 M in #"AgNO"_3# is?

1 Answer
Truong-Son N.
Dec 15, 2017

I got #1.0 xx 10^(-8)"M"#.

The way I see this is that the #"KCN"# and #"AgNO"_3# react to form an initial concentration of #"Ag"("CN")_2^(-)#. The formation of #"Ag"("CN")_2^(-)# is extremely favorable; we know that

#K_f = 1/(4.0 xx 10^(-10)) = 2.5 xx 10^9#.

So, this is first a limiting reactant problem, and then it becomes a common-ion effect problem with leftover excess reactant.

We assume that these solutions are those concentrations after they were mixed, so that the solution volume can be arbitrarily #"1 L"#.

#2"KCN"(aq) + "AgNO"_3(aq) -> "K"["Ag"("CN")_2] (aq) + "KNO"_3(aq)#

I assume you can realize that #"AgNO"_3# is the limiting reactant. We need twice as much #"KCN"#, so we use #"0.08 M"# of the #"KCN"# to convert all #"0.04 M"# of the #"Ag"^(+)# to #"Ag"("CN")_2^(-)# at first, and are left with #"0.04 M KCN"#.

So, our ICE table begins with #"0.04 M Ag"("CN")_2^(-)# and #"0.04 M CN"^(-)#.

#"Ag"("CN")_2^(-)(aq) rightleftharpoons "Ag"^(+)(aq) + 2"CN"^(-)(aq)#

#"I"" ""0.04 M"" "" "" "" ""0.00 M"" "" ""0.04 M"#
#"C"" "-x" "" "" "" "" "+x" "" "" "+2x#
#"E"" "(0.04 - x)"M"" "" "x" M"" "" "(0.04 + 2x)" M"#

This gives an equilibrium expression of:

#K_D -= K_f^(-1)#

#= 4.0 xx 10^(-10) = (["Ag"^(+)]["CN"^(-)]^2)/(["Ag"("CN")_2^(-)])#

#= (x(0.04 + 2x)^2)/(0.04 - x)#

When I use the small x approximation here, I get:

#4.0 xx 10^(-10) = (x(0.04)^2)/(0.04)#

#= 0.04x#

#=> color(blue)(x = ["Ag"^(+)]) = K_D/0.04 = color(blue)(1.0 xx 10^(-8) "M")#

Related questions
Impact of this question
13013 views around the world
You can reuse this answer
Creative Commons License