Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is the concentration of Cu2+ ions in the solution after the addition of NH3? The Kf for Cu(NH3)42+ is 1.7x1013.

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-03-29T13:30:28+02:00

Answer:

[tex]4.6*10^{-14}[/tex] M

Explanation:

Concentration of [tex]Cu^{2+} = [Cu(NO_3)_2][/tex] = 0.020 M

Constructing an ICE table;we have:

[tex]Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}[/tex]

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: [tex]K_f =1.7*10^{13}[/tex]

[tex]K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}[/tex]

[tex]1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}[/tex]

Since x is so small; 0.40 -4x = 0.40

Then:

[tex]1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}[/tex]

[tex]1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}[/tex]

[tex]1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x[/tex]

[tex]8.704*10^9-4.352*10^{11}x =x[/tex]

[tex]8.704*10^9 = 4.352*10^{11}x[/tex]

[tex]x = \frac{8.704*10^9}{4.352*10^{11}}[/tex]

[tex]x = 0.0199999999999540[/tex]

[tex]Cu^{2+}= 0.020 - 0.019999999999954[/tex]

[tex]Cu^{2+} = 4.6*10^{-14}[/tex] M