The tetramminecopper(ii) ion, cu(nh3)4 2+, has a formation constant kf = 1.1 x 10^13. what is the minimum concentration of free ammonia in solution required to ensure that at least 99.9% of the dissolved copper(ii) ion is found in the form of its ammonia complex? (a) 9 x 10^–14 m (b) 9 x 10^–11 m (c) 8 x 10^–4 m (d) 3 x 10^–3 m.

the answer is d. show how you get to the solution

The correct answer is option D, that is, 3 × 10⁻³ M.  

The following is the equation that represent the equilibrium for the formation of tetramminecopper (II) ion:  

Kf = [Cu(NH₃)₄)] / [Cu] [NH₃]₄ Equation (i)

It is known that 99.9 % of copper is in the form of ammonia complex and 0.01 % of it is found in its free state, the reaction will be:  

[Cu(NH₃)₄] / [Cu] = 0.999 / 0.001

Substituting the equation (i) we get:  

1.1 × 10¹³ = 0.999 / (0.001) (NH₃)₄

After solving for ammonia we get:  

[NH₃] = (999 / 1.1 × 10¹³)1/4

[NH₃] = 3.08 × 10⁻³ M

explanation: no but i do not usually comment

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole