# CoreChem:The Common-Ion Effect

### From ChemPRIME

Suppose we have a saturated solution of lead chloride in equilibrium with the solid salt:

If we increase the chloride-ion concentration, 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 product can be used to calculate how much the lead-ion concentration is decreased by the common-ion effect. Suppose we mix 10 cm^{3} of a saturated solution of lead chloride with 10 cm^{3} of concentrated hydrochloric acid (12 *M* HCl). Because of the twofold dilution, the chloride-ion concentration in the mixture will be 6 mol dm^{–3}. Feeding this value into equation 7 from the solubility product section, we then have the result

or

so that

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(NO_{3})_{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 product and defined by

(1)

has a value which is less than the solubility product 1.70 × 10^{–5} mol^{3} 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 *K*_{sp}, corresponds to concentrations which are too large for the solution to be at equilibrium. When this is the case, precipitation 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

precipitation occurs

while if no precipitation occurs

**EXAMPLE 1** Decide whether CaSO_{4} will precipitate or not when (a)100 cm^{3} of 0.02 *M* CaCl_{2} and 100 cm³ of 0.02 *M* Na_{2}SO_{4} are mixed, and also when (b) 100 cm^{3} of 0.002 *M* CaCl_{2} and 100 cm³ of 0.002 *M* Na_{2}SO_{4} are mixed. *K*_{sp} = 2.4 × 10^{–5} mol^{2} dm^{–6}.

**Solution**

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

so that the ion-product *Q* is given by

or

Since *Q* is larger than *K*_{sp}(2.4 × 10^{–5} mol^{2} dm^{–6}), precipitation will occur.

**b)** In the second case

and

Since *Q* is now less than *K*_{sp}, no precipitation will occur.

**EXAMPLE 2** Calculate the mass of CaSO_{4} precipitated when 100 cm^{3} of 0.0200 *M* CaCl_{2} and 100 cm^{3} of 0.0200 *M* Na_{2}SO_{4} 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 cm^{3} of 0.02 *M* CaCl_{2} was used, we have

similarly

If we now indicate the amount of CaSO_{4} precipitated as *x* mmoles, we can set up a table in the usual way:

Species
| Ca
^{2+} (aq) | SO
_{4}^{2–} (aq) |

Initial amount (mmol)
| ||

Amount reacted (mmol)
| ||

Equilibrium amount (mmol)
| ||

Equilibrium concentration (mmol cm
^{–3}) |

Thus

or

Rearranging,

or

so that

Since 1.020 mmol CaSO_{4} is precipitated, the mass precipitated is given by

*Note*: Because the solutions are so dilute and because CaSO_{4} has a fairly large solubility product, only about half (1.02 mmol out of a total of 2.00 mmol) the Ca^{2+} ions are precipitated. If we wished to determine the concentration of Ca^{2+} ions in tap water or river water, where it is quite low, it would be foolish to try to precipitate the Ca as CaSO_{4}. Another method would have to be found.