chapter regarding Ksp and complex ions Our system is 0.020 moles solid CuCl 2 ad

Question

chapter regarding Ksp and complex ions

Our system is 0.020 moles solid CuCl2 added to 1L of 0.100M NaCN. We are ultimately trying to calculate the Cu2+ ion concentration.
There are 3 reactions happening in this system simultaneously. One is the dissolving of copper cholride (CuCl2)- Ksp= 1.0 x10-6, the second is the formation of the copper cyanide (CuCN) complex Kf =1.0 x1025 and the third is the reduction of CN- concentration due to formation of HCN Kb= 1.6 x10 -5

A) What happens to copper chloride. Write out a new equilibrium expression for the reaction of copper chloride with cyanide ion. Find the value of K for equilibrium.

B) Assume that it did all dissolve and now we have that 0.020mol of CuCl2 dissolved in our solution. We can assume that essentially all the copper is complexed to Cu(CN)42-. Use this information to calculate the [CN-] and [Cu(CN)42-] after all the copper chloride dissolves but before any other reactions happen. (hint: stoichiometry)

Explanation / Answer

A) Copper (II) chloride dissolves in the NaCN solution to form a tetracyanocopper(II) complex as per the below equations:

CuCl2 (s) <========> Cu2+ (aq) + 2 Cl- (aq) ;                            Ksp = 1.0*10-6

Cu2+ (aq) + 4 CN- (aq) <=======> [Cu(CN)4]2- (aq);                 Kf = 1.0*1025

Adding,

CuCl2 (s) + 4 CN- (aq) <=======> [Cu(CN)4]2- (aq) + 2 Cl- ;     K = Ksp*Kf

                                                                                                            = (1.0*10-6)*(1.0*1025)

                                                                                                            = 1.0*1019

The equilibrium constant for the formation of tetracyanocopper(II) complex is 1.0*1019 (ans).

B) Now, for this part of the problem, we first calculate the molar concentration of CuCl2 in 1.0 L solution as (0.020 mole/1.00 L) = 0.020 M.

Also note [CuCl2] = [Cu2+] = 0.020 M (as there is a 1:1 ratio as per the first equation); One point that has already been mentioned is that we need to consider all the Cu2+ is complexed with CN- form the complex.

We write down the reaction as before

Cu2+ (aq) + 4 CN- (aq) <=======> [Cu(CN)4]2- (aq)

initial                0.020 M          1.00 M                                  0

change           - 0.020 M    - 4*(0.020) M                     + 0.020 M

final                        0                   0.92 M                            0.020 M

The final concentration is not the same as the equilibrium concentration; infact the final concentration is the initial concentration while evaluating the equilibrium constant. We shall calculate the equilibrium concentration by setting up an ICE chart. Also, to calculate the concentration of free CN- in the solution, we need to consider dissociation of the tetracyanocopper(II) complex. The dissociation constant is given as

Kd = 1/Kf = 1/(1.0*1025) = 1.0*10-25

[Cu(CN)4]2- (aq) <=======> Cu2+ (aq) + 4 CN- (aq)

initial                    0.020 M                                    0                 0.92 M

change                    - x                                         + x                 + 4x

equilibrium         (0.020 – x)                                 x              (0.92 + 4x)

Now, the dissociation constant is

Kd = [Cu2+][CN-]4/[Cu(CN)4]2- = (x)(0.92 + 4x)4/(0.020 – x)

We need to assume that the dissociation of the complex is extremely small, hence x is small compared to 0.020 M or 0.92 M.

Therefore,

1.0*10-25 = x.(0.92)4/(0.020) = 0716x/0.020 = 35.82x

or, x = 2.79*10-27 2.80*10-27

The molar concentration of [Cu(CN)4]2- is 2.80*10-27 M and that of free CN- is (0.92 + 4*2.80*10-27) M = 0.92 M (ans)

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