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 1 of 4: Acidity of hydrated Cu²⁺

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Intuition

Because the dissociated fraction is tiny ( $\sim 0.2$ %), the initial Cu²⁺ concentration can be used undiluted in the denominator.

[\ce{Cu(H2O)4}]^{2+} + \ce{H2O} \rightleftharpoons \ce{H3O+} + [\ce{Cu(OH)(H2O)3}]^{+};\qquad K_a = \frac{(10^{-4.65})^2}{10^{-2}} = 5.0\times10^{-8},\quad pK_a = 7.30

Analysis

The hydrated ion donates a proton to water. With $[\ce{H3O+}] = 10^{-4.65} = 2.24\times10^{-5}$ and $[\ce{Cu(OH)(H2O)3+}] = [\ce{H3O+}]$ , $K_a = (2.24\times10^{-5})^2/1.00\times10^{-2} = 5.0\times10^{-8}$ , hence p $K_a$ = 7.30.