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 4 of 4: Equilibrium composition

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2\times10^{6}[\ce{Cu+}]^2 + [\ce{Cu+}] - 10^{-2} = 0\ \Rightarrow\ [\ce{Cu+}] = 7.1\times10^{-5},\quad [\ce{Cu^{2+}}] = 5.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}

Analysis

Mass balance gives $[\ce{Cu+}] + 2[\ce{Cu^{2+}}] = 10^{-2}$ (each Cu²⁺ formed consumes two Cu⁺). Combining with $[\ce{Cu^{2+}}] = 10^{6}[\ce{Cu+}]^2$ yields the quadratic equation; solving gives $[\ce{Cu+}] = 7.07\times10^{-5}$ M and $[\ce{Cu^{2+}}] = 4.96\times10^{-3}$ M, so virtually all copper(I) disproportionates.