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 2 of 4: Precipitation of Cu(OH)₂

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[\ce{OH-}] = \sqrt{\frac{K_{sp}}{[\ce{Cu^{2+}}]}} = \sqrt{\frac{10^{-20}}{10^{-2}}} = 10^{-9}\ \mathrm{mol\,dm^{-3}}\ \Rightarrow\ \mathrm{pH} = 5.00

Analysis

Precipitation starts when $[\ce{Cu^{2+}}][\ce{OH-}]^2 = K_{sp}$ , i.e. $[\ce{OH-}] = 10^{-9}$ M, pOH = 9, pH = 5. At this pH $[\ce{Cu(OH)(H2O)3+}]/[\ce{Cu^{2+}}] = K_a/[\ce{H+}] = 10^{-7.3}/10^{-5} \approx 1/200$ , so the conjugate base is indeed negligible.