Problem 2

Aqueous solutions of copper salts. The pH of a

1.00\times10^{-2}

mol dm

^{-3}

copper(II) nitrate solution is 4.65. (2.1) Write the equation for the formation of the conjugate base of the hydrated

\ce{Cu^{2+}}

ion. (2.2) Calculate the p

K_a

of the corresponding acid–base pair. (2.3) Given

K_{sp}(\ce{Cu(OH)2}) = 1\times10^{-20}

, at what pH does

\ce{Cu(OH)2}

start to precipitate? (2.4) Using

E^0(\ce{Cu+/Cu}) = +0.52

V and

E^0(\ce{Cu^{2+}/Cu+}) = +0.16

V, write the disproportionation of

\ce{Cu+}

and calculate its equilibrium constant. (2.5) Calculate the composition (mol dm

^{-3}

) of the solution obtained by dissolving

1.00\times10^{-2}

mol of copper(I) in

1.0\ \mathrm{dm^3}

of water.

Step 3 of 4: Disproportionation constant

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2\ce{Cu+} \rightleftharpoons \ce{Cu^{2+}} + \ce{Cu};\qquad \log K = \frac{E^0(\ce{Cu+/Cu}) - E^0(\ce{Cu^{2+}/Cu+})}{0.059} = \frac{0.36}{0.059}\ \Rightarrow\ K \approx 10^{6}

Analysis

Disproportionation combines $\ce{Cu+} + e^- \rightarrow \ce{Cu}$ (cathode, $E^0 = 0.52$ V) and $\ce{Cu+} \rightarrow \ce{Cu^{2+}} + e^-$ (anode). $E^0_{\mathrm{cell}} = 0.52 - 0.16 = 0.36$ V, and for $n = 1$ , $\log K = E^0/0.059$ , so $K = [\ce{Cu^{2+}}]/[\ce{Cu+}]^2 \approx 10^{6}$ .

Common pitfall. A positive

E^0

for both couples does not prevent disproportionation; what matters is that

E^0(\ce{Cu+/Cu}) > E^0(\ce{Cu^{2+}/Cu+})