# The Common-Ion Effect

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:57

Suppose we have a saturatedDescribes 1) a solution that contains the equilibrium concentration of a solute, or 2) an organic compound that contains no double or triple bonds (such as an alkane). solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. of lead chloride in equilibriumA state in which no net change is occurring, that is, in which the concentrations of reactants and products remain constant; chemical equilibrium is characterized by forward and reverse reactions occurring at the same rate. with the solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. saltAn ionic compound that can be formed by replacing the hydrogen ion of an acid with a different cation.:

$\text{PbCl}_{2}({s})\rightleftharpoons \text{Pb}^{2+}({aq}) + \text{2Cl}^{-}({aq})$

If we increase the chloride-ion concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated., Le Chatelier’s principle predicts that the equilibrium will shift to the left. More lead chloride will precipitate, and the concentration of lead ions will decrease. A decrease in concentration obtained in this way is often referred to as the common-ion effect.

The solubility productThe equilibrium constant expression for the dissolution of an electrolyte; the reactant is a solid and its concentration does not appear in the expression, which is a product of the concentrations of the products (raised the to appropriate powers). can be used to calculate how much the lead-ion concentration is decreased by the common-ion effect. Suppose we mix 10 cm3 of a saturated solution of lead chloride with 10 cm3 of concentratedIncreased the concentration of a mixture or solution (verb). Having a large concentration (adjective). hydrochloric acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond. (12 M HCl). Because of the twofold dilutionThe addition of solvent to a solution (or mixture) to decrease the concentration of a solute (or component)., the chloride-ion concentration in the mixtureA combination of two or more substances in which the substances retain their chemical identity. will be 6 mol dm–3. Feeding this value into equation 7 from the solubility product section, we then have the result

${K}_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^{2}\,$

or              $\text{1.70} \times \text{10}^{-5} \text{mol}^{3} \text{dm}^{-9}=[\text{Pb}^{2+}](\text{6 mol dm}^{-3})^{2}$

so that      $[\text{ Pb}^{2+}]=\frac{\text{1.70 }\times \text{ 10}^{-5}\text{ mol}^{3}\text{ dm}^{-9}}{\text{36 mol}^{2}\text{ dm}^{-6}}=\text{4.72 }\times \text{ 10}^{-7}\text{ dm}^{3}$

We have thus lowered the lead-ion concentration from an initial value of 1.62 × 10–2 mol dm–3 (see Example 1 from the section on the solubility product) to a final value of 4.72 × 10–7 mol dm–3, a decrease of about a factor of 30 000! As a result, we have at our disposal a very sensitive test for lead ions. If we mix equal volumes of 12 M HCl and a test solution, and no precipitate occurs, we can be certain that the lead-ion concentration in the test solution is below 2 × 4.72 × 10–7 mol dm–3.

Because it tells us about the conditions under which equilibrium is attained, the solubility product can also tell us about those cases in which equilibrium is not attained. If extremely dilute solutions of Pb(NO3)2 and KCl are mixed, for instance, it may be that the concentrations of lead ions and chloride ions in the resultant mixture are both too low for a precipitate to form. In such a case we would find that the product Q, called the ion productAn equilibrium constant expression for a reaction in which the only products are ions and the reactants are such that their concentrations do not appear in the expression; applied to dissolution and autoionization reactions. and defined by

$Q=\text{(}c_{\text{Pb}^{2+}}\text{)(}c_{\text{Cl}^{-}}\text{)}^{2}$      (1)

has a value which is less than the solubility product 1.70 × 10–5 mol3 dm–9. In order for equilibrium between the ions and a precipitate to be established, either the lead-ion concentration or the chloride-ion concentration or both must be increased until the value of Q is exactly equal to the value of the solubility product. The opposite situation, in which Q is larger than Ksp, corresponds to concentrations which are too large for the solution to be at equilibrium. When this is the case, precipitationThe formation of a solid within a solution, often by the combination of cations and anions to form an insoluble ionic compound. occurs, lowering the concentration of both the lead and chloride ions, until Q is exactly equal to the solubility product.

To determine in the general case whether a precipitate will form, we set up an ion-product expression Q which has the same form as the solubility product, except that the stoichiometric concentrations rather than the equilibrium concentrations are used. Then if

$\text{Q}>{K}_{sp}\,$      precipitation occurs

while if      $\text{Q}<{K}_{sp}\,$      no precipitation occurs

EXAMPLE 1 Decide whether CaSO4 will precipitate or not when (a)100 cm3 of 0.02 M CaCl2 and 100 cm³ of 0.02 M Na2SO4 are mixed, and also when (b) 100 cm3 of 0.002 M CaCl2 and 100 cm³ of 0.002 M Na2SO4 are mixed. Ksp = 2.4 × 10–5 mol2 dm–6.

Solution

a) After mixing, the concentration of each species is halved. We thus have

${c}_{\text{Ca}^{2+}}=\text{0.01 mol dm}^{-3}={c}_{\text{SO}_{4}^{2-}}$

so that the ion-product Q is given by

${Q}={c}_{\text{Ca}^{2+}} \times {c}_{\text{SO}_{4}^{2-}} = \text{0.01 mol dm}^{-3} \times \text{ 0.01 mol dm}^{-3}$

or      ${Q}= \text{10}^{-4} \text{ mol}^{2}\text{dm}^{-6}\,$

Since Q is larger than Ksp(2.4 × 10–5 mol2 dm–6), precipitation will occur.

b) In the second case

${c}_{\text{Ca}^{2+}}= \text{0.001 mol dm}^{-3} = {c}_{\text{SO}_{4}^{2-}}$

and      ${Q}={c}_{\text{Ca}^{2+}}\times {c}_{\text{SO}_{4}^{2-}}= \text{1} \times \text{10}^{-6} \text{mol}^{2} \text{dm}^{-6}$

Since Q is now less than Ksp, no precipitation will occur.

EXAMPLE 2 Calculate the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of CaSO4 precipitated when 100 cm3 of 0.0200 M CaCl2 and 100 cm3 of 0.0200 M Na2SO4 are mixed together.

Solution We have already seen in part a of the previous example that precipitation does actually occur. In order to find how much is precipitated, we must concentrate on the amount of each species. Since 100 cm3 of 0.02 M CaCl2 was used, we have

$n_{\text{Ca}^{2+}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}}\times \text{ 100 cm}^{3}=\text{2.00 mmol}$

similarly      $n_{\text{SO}_{4}^{2-}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}} \times \text{ 100 cm}^{3}=\text{2.00 mmol}$

If we now indicate the amount of CaSO4 precipitated as x mmoles, we can set up a table in the usual way:

 Species Ca2+ (aq) SO42– (aq) Initial amount (mmol) $\text{2.00}\,$ $\text{2.00}\,$ Amount reacted (mmol) $-{x}\,$ $-{x}\,$ Equilibrium   amount (mmol) $(\text{2}-{x})\,$ $(\text{2}-{x})\,$ Equilibrium   concentration (mmol cm–3) $\frac{\text{2}-{x}}{\text{200}}$ $\frac{\text{2}-{x}}{\text{200}}$

Thus

${K}_{sp} = [\text{Ca}^{2+}][\text{SO}_{4}^{2-}]\,$

or      $\text{2.4 }\times \text{ 10}^{-5}\text{ mol}^{2}\text{ dm}^{-6}=\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)$

Rearranging,

$\text{200}^{2}\times\text{2.4}\times\text{10}^{-5}=\text{0.96}=(\text{2}-{x})^{2}$

or              $\text{2}-{x}=\sqrt{\text{0}\text{.96}}=\text{0.980}$

so that      ${x}=\text{2}-\text{0.980}=\text{1.020}\,$

Since 1.020 mmol CaSO4 is precipitated, the mass precipitated is given by

\begin{align}m_{\text{CaSO}_{4}}&=\text{1.020 mmol }\times \text{ 136.12 }\frac{\text{mg}}{\text{mmol}}\ \text{ }&=\text{138.9 mg}=\text{0.139 g}\end{align}

Note: Because the solutions are so dilute and because CaSO4 has a fairly large solubility product, only about half (1.02 mmol out of a total of 2.00 mmol) the Ca2+ ions are precipitated. If we wished to determine the concentration of Ca2+ ions in tap water or river water, where it is quite low, it would be foolish to try to precipitate the Ca as CaSO4. Another method would have to be found.