IChO 1977 · Theory problem 5

Solubility of two copper(I) salts in a buffer

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Two copper(I) salts of the organic acids \(\ce{HA}\) and \(\ce{HB}\), slightly soluble in water, form a saturated solution in buffer of a given pH.

Problems:

5.1 What will be the concentration of \(\ce{Cu+}\) cations in the solution if the solubility products of the two salts are \(K_{\mathrm{s}}(\ce{CuA})\) and \(K_{\mathrm{s}}(\ce{CuB})\) and the ionisation constants of the acids are \(K_{\mathrm{a}}(\ce{HA})\) and \(K_{\mathrm{a}}(\ce{HB})\)?

Official English text of the problem.

Answers

  • 5.1 \[ \boxed{ [\ce{Cu+}] = \sqrt{ K_{\mathrm{s}}(\ce{CuA}) \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HA})}\right) + K_{\mathrm{s}}(\ce{CuB}) \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HB})}\right) }} \] Here \(h=[\ce{H+}]=10^{-\mathrm{pH}}\ \mathrm{mol\,L^{-1}}\) in the ideal-solution approximation, with equilibrium constants expressed in concentration form.

Solution

5.1

Let \(c=[\ce{Cu+}]\), the concentration of the copper(I) ion ([Cu+]), and \(h=[\ce{H+}]\), the concentration of the hydrogen ion, or proton ([H+]).

The structures of the organic acids \(\ce{HA}\) and \(\ce{HB}\), their conjugate-base anions \(\ce{A-}\) and \(\ce{B-}\), and their salts are unspecified, so unique names and SMILES cannot be assigned to them.

We use the intended equilibrium model: the solution is saturated with respect to both salts, the buffer fixes the pH without supplying either acid residue, and copper remains as free \(\ce{Cu+}\).

The dissolution equilibria are \[ \ce{CuA(s) <=> Cu+(aq) + A-(aq)}, \qquad \ce{CuB(s) <=> Cu+(aq) + B-(aq)}. \] Consequently, \[ K_{\mathrm{s}}(\ce{CuA})=c[\ce{A-}], \qquad K_{\mathrm{s}}(\ce{CuB})=c[\ce{B-}], \] and hence \[ [\ce{A-}]=\frac{K_{\mathrm{s}}(\ce{CuA})}{c}, \qquad [\ce{B-}]=\frac{K_{\mathrm{s}}(\ce{CuB})}{c}. \]

The acid dissociation equilibria are \[ \ce{HA(aq) <=> H+(aq) + A-(aq)}, \qquad \ce{HB(aq) <=> H+(aq) + B-(aq)}. \] Their constants give \[ [\ce{HA}]=\frac{h[\ce{A-}]}{K_{\mathrm{a}}(\ce{HA})}, \qquad [\ce{HB}]=\frac{h[\ce{B-}]}{K_{\mathrm{a}}(\ce{HB})}. \]

Every mole of either salt dissolved supplies one mole of copper and one mole of the corresponding acid residue. Protonation preserves the total amount of each residue. Thus the material balance is \[ c=[\ce{A-}]+[\ce{HA}]+[\ce{B-}]+[\ce{HB}]. \] Substituting the equilibrium expressions, \[ c= \frac{K_{\mathrm{s}}(\ce{CuA})}{c} \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HA})}\right) + \frac{K_{\mathrm{s}}(\ce{CuB})}{c} \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HB})}\right). \] Multiplying by \(c\) and taking the positive square root yields \[ \boxed{ [\ce{Cu+}] = \left[ K_{\mathrm{s}}(\ce{CuA}) \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HA})}\right) + K_{\mathrm{s}}(\ce{CuB}) \left(1+\frac{h}{K_{\mathrm{a}}(\ce{HB})}\right) \right]^{1/2} }. \]

In concentration form, \(h/K_{\mathrm{a}}\) is dimensionless and each \(K_{\mathrm{s}}\) has units \(\mathrm{mol^2\,L^{-2}}\), giving the required concentration in \(\mathrm{mol\,L^{-1}}\). No numerical values are supplied.

The solution was checked against the official solution.