A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is

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A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is

the concentration of cu2+(aq ) in the solution?

Chemistry

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castortr0y [923] · Answer 1 · 2026-03-05T00:55:20+03:00

<span>Response: A 1.00 L solution that includes 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). Contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine. Using the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with double that amount of en = 0.000600 mol of en. Thus, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted leaves 0.00180 mol en unreacted. According to the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to yield an equivalent of 0.000300 moles of Cu(en)2^2+ The formation constant Kf for Cu(en)2^2+ is 1x10^20. Therefore, 1 Cu+2 and 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20 = [0.000300] / [Cu+2] [0.00180 ]^2 Solving for [Cu+2] gives [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Thus, Cu+2 = 9.26 e-19 Molar. Since Kf only has 1 significant figure, round that to 9 X 10^-19 Molar Cu+2.</span>