Question

09 Question (1 point) A 1.00-L solution contains 2.25\times10^{-4} M Cu(NO$_3$)$_2$ and 1.00\times10$^{-3}$ M ethylenediamine (en). The $K_f$ for Cu(en)$_2^{2+}$ is 1.00 \times 10$^{20}$. Cu$^{2+}$(aq) + 2en(aq) $\rightleftharpoons$ Cu(en)$_2^{2+}$(aq) $\frac{[Cu(en)_2^{2+}]}{[Cu^{2+}][en]^2}$

          09 Question (1 point)
A 1.00-L solution contains 2.25\times10^{-4} M Cu(NO$_3$)$_2$ and 1.00\times10$^{-3}$ M ethylenediamine (en). The $K_f$ for Cu(en)$_2^{2+}$ is 1.00 \times 10$^{20}$.
Cu$^{2+}$(aq) + 2en(aq) \(\rightleftharpoons\) Cu(en)$_2^{2+}$(aq)
$\frac{[Cu(en)_2^{2+}]}{[Cu^{2+}][en]^2}$

09 Question (1 point)
A 1.00-L solution contains 2.25×10^-4 M Cu(NO3)2 and 1.00×10^-3 M ethylenediamine (en). The Kf for Cu(en)2^2+ is 1.00 ×10^20.
Cu^2+(aq) + 2en(aq) ⇌ Cu(en)2^2+(aq)
([Cu(en)2^2+])/([Cu^2+][en]^2)

Added by Jillian M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Preeti Kumari

03/23/2022

Step 1

25 x 10^(-4) M [en] = 1.00 x 10^(-3) M Show more…

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09 Question (1 point) A 1.00-L solution contains 2.25 imes 10^(-4)MCu(NO_(3))_(2) and 1.00 imes 10^(-3)M ethylenediamine (en). The K_(f) for Cu(en)_(2)^(2+) is 1.00 imes 10^(20). Cu^(2+)(aq)+2en(aq)⇌Cu(en)_(2)^(2+)(aq) K_(f)=(|_(_())Cu(en)_(2)^(2+)_(_())|)/([Cu^(2+)][en]^(2)) 09 Question (1point) A 1.00-L solution contains 2.2510-4 M Cu(NO3)2 and 1.0010-3 M ethylenediamine (en). The Kf for Cu(en)2* is 1.00 1020. See page 795 Cu2+(aq)+2en(aq) Cu(en)3+(aq) [cu(en)2+] Kf= [cu2+][en]2