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A solution is formed by mixing 50.0 mL of 10.0 M NaX with
Chapter 16, Problem 67(choose chapter or problem)
A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of \(2.0\times10^{-3}\ M\ \mathrm{CuNO}_3\). Assume that Cu(I) forms complex ions with \(\mathrm{X}^{-}\) as follows:
\(\begin{aligned}\mathrm{Cu}^+(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}(aq)&\ \ K_1=1.0\times10^2\\ \mathrm{CuX}^-(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}_2^{\ -}(aq)&K_2=1.0\times10^4\\ \mathrm{CuX}_2^{\ -}(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}_3^{\ 2-}(aq)&K_3=1.0\times10^3\end{aligned}\)
with an overall reaction
\(\mathrm{Cu}^+(aq)+3\mathrm{X}^-(aq)\ \leftrightharpoons\ \mathrm{CuX}_3^{\ 2-}(aq)\quad\ \ \ K=1.0\times10^9\)
Calculate the following concentrations at equilibrium.
a. \(\mathrm{CuX}_3^{\ 2-}\)
b. \(\mathrm{CuX}_2^{\ -}\)
c. \(\mathrm{Cu}^{+}\)
Questions & Answers
QUESTION:
A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of \(2.0\times10^{-3}\ M\ \mathrm{CuNO}_3\). Assume that Cu(I) forms complex ions with \(\mathrm{X}^{-}\) as follows:
\(\begin{aligned}\mathrm{Cu}^+(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}(aq)&\ \ K_1=1.0\times10^2\\ \mathrm{CuX}^-(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}_2^{\ -}(aq)&K_2=1.0\times10^4\\ \mathrm{CuX}_2^{\ -}(aq)+\mathrm{X}^-(aq)&\leftrightharpoons\mathrm{\ CuX}_3^{\ 2-}(aq)&K_3=1.0\times10^3\end{aligned}\)
with an overall reaction
\(\mathrm{Cu}^+(aq)+3\mathrm{X}^-(aq)\ \leftrightharpoons\ \mathrm{CuX}_3^{\ 2-}(aq)\quad\ \ \ K=1.0\times10^9\)
Calculate the following concentrations at equilibrium.
a. \(\mathrm{CuX}_3^{\ 2-}\)
b. \(\mathrm{CuX}_2^{\ -}\)
c. \(\mathrm{Cu}^{+}\)
ANSWER:Step 1 of 5
The concentration of the ligand and metal ion in the mixed solution before any reaction occurs is as.
\(\begin{aligned}\left[\mathrm{Cu}^+\right]&=\frac{(50\mathrm{ml})\left(2.0\times10^{-3}\mathrm{\ m}\right)}{(50\ \mathrm{ml}+50\ \mathrm{ml})}=1.0\times10^{-3}\mathrm{M}\\ \left[\mathrm{X}^-\right]&=\frac{(50\mathrm{\ ml})(10.0\mathrm{\ m})}{(50\mathrm{\ ml}+50\mathrm{\ ml})}=5.0\mathrm{M}\end{aligned}\)
The overall reaction is
\(\mathrm{Cu}^++3\mathrm{X}^-\rightarrow\mathrm{Cu}\mathrm{X}_3^{-2};\ \mathrm{K}=1.0\times10^9\)