A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium:
-[Cu(CN) ₂^-]

Question

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows:

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10<sup>-3</sup> M CuNO<sub>3</sub>. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) <sub>2</sub><sup>-</sup>] A) 1.0 × 10<sup>-3</sup> M B) 4.0 × 10<sup>-12</sup> M C) 2.0 × 10<sup>-7</sup> M D) 2.0 × 10<sup>-3</sup> M E) none of these

Calculate the following concentrations at equilibrium:
-[Cu(CN) ₂^-]

Quizplus · Accepted Answer

First, let's write out the balanced equation for the reaction:

Cu+ + CN- --> Cu(CN)2-

We start with 50.0 mL of 10.00 M NaCN, which contains 0.50 moles of CN-. We also start with 50.0 mL of 2.0 × 10^-3 M Cu+, which contains 1.0 × 10^-4 moles of Cu+.

The reaction between Cu+ and CN- will consume some of the CN-, forming the Cu(CN)2- complex ion. Let's define x as the amount of CN- that reacts to form the complex. Then:

Cu+ + x CN- --> Cu(CN)2-

We can set up an ICE (initial, change, equilibrium) table:

Cu+         CN-          Cu(CN)2- 
I        1.0 × 10^-4   0.50          0 
C       -x            -x            +x 
E        1.0 × 10^-4 - x  0.50 - x   x

We can use the Kf expression to solve for x:

Kf = [Cu(CN)2-]/([Cu+][CN-]^2)

Kf = 4.8 × 10^20 M^-2 from the given data

Kf = x/([1.0 × 10^-4][0.50 - x]^2)

Simplifying:

4.8 × 10^20 = x/(1.25 × 10^-5 - 1.5x + x^2)

We can assume that x is small compared to 0.50, so we can simplify further:

4.8 × 10^20 = x/(1.25 × 10^-5)

x = 6.0 × 10^15 M

Now we can calculate the concentration of Cu(CN)2- at equilibrium:

[Cu(CN)2-] = x/0.05 L = 1.2 × 10^-13 M

(Note that the volume of the final solution is 100.0 mL.)

So the answer is (C) 2.0 × 10^-7 M.

A Solution Is Formed by Mixing 50