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14.6 THE SOLUBILITY PRODUCT

In Sec. 11.1we saw that there are some salts which dissolve in water to only a very limited extent. For example, if BaSO₄ crystals are shaken with water, so little dissolves that it is impossible to see that anything has happened. Nevertheless, the few Ba²⁺(aq) and SO₄^2–(aq) ions that do go into solution increase the conductivity of the water, allowing us to measure their concentration. We find that at 25°C

[Ba²⁺] = 0.97 × 10^–5 mol dm^–3 = [SO₄^2–] (14.26)

so that we would describe the solubility of BaSO₄ as 0.97 × 10^–5 mol dm^–3 at this temperature. The solid salt and its ions are in dynamic equilibrium, and so we can write the equation

BaSO₄(s) $\rightleftharpoons$ Ba²⁺(aq) + SO₄^2–(aq) (14.27)

As in other dynamic equilibria we have discussed, a particular Ba²⁺ ion will sometimes find itself part of a crystal and at other times find itself hydrated and in solution.

Since the concentration of BaSO₄ has a constant value, it can be incorporated into K_c for Eq. (14.27). This gives a special equilibrium constant called the solubility product K_sp:

K_sp = K_c[BaSO₄] = [Ba²⁺][ SO₄^2–] (14.28)

For BaSO₄, K_sp is easily calculated from the solubility by substituting Eq. (14.26) into (14.28):

K_sp = (0.97 × 10^–5 mol dm^–3)( 0.97 × 10^–5 mol dm^–3) = 0.94 × 10^–10 mol² dm^–6

TABLE 14.6 The Solubility Product K_sp for Some Sparingly Soluble Salts at 25°C.

Compound	K_sp	Compound	K_sp
BaF₂	2.4 × 10^–5 mol³ dm^–9	PbS	7 × 10^–29 mol² dm^–6
BaSO₄	1.0 × 10^–10 mol² dm^–6	Mg(OH)₂	8.9 × 10^–12 mol³ dm^–9
CaCO₃	4.7 × 10^–9 mol² dm^–6	HgS	1.6 × 10^–54 mol² dm^–6
CaF₂	3.9 × 10^–11 mol³ dm^–9	NiS	3 × 10^–21 mol² dm^–6
CaSO₄	2.4 × 10^–5 mol² dm^–6	AgC₂H₃O₂ *	2.3 × 10^–3 mol² dm^–6
Fe(OH)₂	1.8 × 10^–15 mol³ dm^–9	AgCl	1.7 × 10^–10 mol² dm^–6
Fe(OH)₃	6 × 10^–38 mol⁴ dm^–12	AgBr	5.0 × 10^–13 mol² dm^–6
FeS	4 × 10^–19 mol³ dm^–9	AgI	8.5 × 10^–17 mol² dm^–6
PbCl₂	1.7 × 10^–5 mol³ dm^–9	Ag₂CrO₄	1.9 × 10^–12 mol³ dm^–9
PbSO₄	1.3 × 10^–8 mol² dm^–6	ZnS	2.5 × 10^–22 mol² dm^–6

* Silver acetate

In the general case of an ionic compound whose formula is A_xB_y, the equilibrium can be written

A_xB_y(s) $\rightleftharpoons$ xA^m+(aq) + yBⁿ⁺(aq) (14.29)

The solubility product is then

K_sp = [A^m+]^x[ Bⁿ⁺]^y (14.30)

Solubility products for some of the more common sparingly soluble compounds are given in Table 14.6.

EXAMPLE 14.17 When crystals of PbCl₂ are shaken with water at 25°C, it is found that 1.62 × 10^–2 mol PbCl₂ dissolves per cubic decimeter of solution. Find the value of K_sp at this temperature.

Solution We first write out the equation for the equilibrium:

PbCl₂(s) $\rightleftharpoons$ Pb²⁺(aq) + 2Cl^–(aq)...(14.31)

so that

K_sp(PbCl₂) = [Pb²⁺][Cl^–]²...(14.32)

Since 1.62 × 10^–2 mol PbCl₂ dissolves per cubic decimeter, we have

[Pb²⁺] = 1.62 × 10^–2 mol dm^–3

while [Cl^–] = 2 × 1.62 × 10^–2 mol dm^–3

since 2 mol Cl^– ions are produced for each mol PbCl₂ which dissolves. Thus

K_sp = (1.62 × 10^–2 mol dm^–3)( 2 × 1.62 × 10^–2 mol dm^–3)² = 1.70 × 10^–5 mol³ dm^–9

EXAMPLE 14.18 The solubility product of silver chromate, Ag₂CrO, is 1.0 × 10^–12 mol³ dm^–9. Find the solubility of this salt.

Solution Again we start by writing the equation

Ag₂CrO(s) [[Image:]] 2Ag²⁺(aq) + CrO₄^2–(aq)

from which

K_sp = [Ag⁺]² [CrO₄^2–] = 1.0 × 10^–12 mol³ dm^–9

Let the solubility be x mol dm^–3. Then

[CrO₄^2–] = x mol dm^–3 and [Ag⁺] = 2x mol dm^–3

Thus K_sp = (2x mol dm^–3)² x mol dm^–3 = (2x)² x mol³ dm^–9 = 1.0 × 10^–12 mol³ dm^–9

or 4x³ = 1.0 × 10^–12

and $x^{\text{3}}=\frac{\text{1}\text{.0}}{\text{4}}\text{ }\times \text{ 10}^{-\text{12}}=\text{2}\text{.5 }\times \text{ 10}^{-\text{13}}=\text{250 }\times \text{ 10}^{-\text{15}}$

so that $x=\sqrt[\text{3}]{\text{250}}\text{ }\times \text{ }\sqrt[\text{3}]{\text{10}^{-\text{15}}}=\text{6}\text{.30 }\times \text{ 10}^{-\text{5}}$

Thus the solubility is 6.30 × 10^–5 mol dm^–3.

Note: If your calculator does not have a $\sqrt[y]{x}$ or a x^y button, a cube root can be found from the relationship $\sqrt[3]{x}$ = x^⅓ = antilog (⅓ log x).If your calculator cannot find a log, it is not difficult to find cube root of 250 by trial and error. Since 6³ = 216 and 7³ = 343, the cube root is between 6 and 7, probably closer to 6. Continuing in this fashion we quickly find that x is very close to 6.3.

The Common-Ion Effect

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

PbCl₂(s) $\rightleftharpoons$ Pb²⁺(aq) + 2Cl^–(aq)

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 cm³ of a saturated solution of lead chloride with 10 cm³ 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 Eq. (14.32), we then have the result

K_sp = [Pb²⁺][ Cl^–]² 1.0

or 1.70 × 10^–5 mol³ dm^–9 = [Pb²⁺]( 6 mol dm^–3)²

so that $\text{ }\!\![\!\!\text{ Pb}^{\text{2+}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1}\text{.70 }\times \text{ 10}^{-\text{15}}\text{ mol}^{\text{3}}\text{ dm}^{-\text{9}}}{\text{36 mol}^{\text{2}}\text{ dm}^{-\text{6}}}=\text{4}\text{.72 }\times \text{ 10}^{-\text{7}}\text{ dm}^{-\text{3}}$

We have thus lowered the lead-ion concentration from an initial value of 1.62 × 10^–2 mol dm^–3 (see Example 14.17) 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 KC1 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

$Q=\text{(}c_{\text{Pb}^{\text{2+}}}\text{)(}c_{\text{Cl}^{-}}\text{)}^{\text{2}}$ (14.33)

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 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

Q > K_sp precipitation occurs

while if Q < K_sp no precipitation occurs

EXAMPLE 14.19 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. K_sp = 2.4 × 10^–5 mol² dm^–6.

Solution

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

c_Ca²⁺ = 0.01 mol dm^–3 = c_SO₄^2–

so that the ion-product Q is given by

Q = c_Ca²⁺ × c_SO₄^2– = 0.01 mol dm^–3 ×0.01 mol dm^–3

or Q = 10^–4 mol² dm^–6

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

b) In the second case

c_Ca²⁺ = 0.001 mol dm^–3 = c_SO₄^2–

and Q = c_Ca²⁺ × c_SO₄^2– = 1 × 10^–6 mol² dm^–6

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

EXAMPLE 14.20 Calculate the mass 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

$c_{\text{Ca}^{\text{2+}}}=\text{0}\text{.0200 }\frac{\text{mmol}}{\text{cm}^{\text{3}}}\text{ }\times \text{ 100 cm}^{\text{3}}=\text{2}\text{.00 mmol}$

similarly $c_{\text{SO}_{\text{4}}^{\text{2}-}}=\text{0}\text{.0200 }\frac{\text{mmol}}{\text{cm}^{\text{3}}}\text{ }\times \text{ 100 cm}^{\text{3}}=\text{2}\text{.00 mmol}$

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

Thus

K_sp = [Ca²⁺][ SO₄^2–]

or $\text{2}\text{.4 }\times \text{ 10}^{-\text{5}}\text{ mol}^{2}\text{ dm}^{-\text{6}}=\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-\text{3}} \right)\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-\text{3}} \right)$

Rearranging,

200² × 2.4 × 10^–5 = 0.96 = (2 – x)²

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

so that x = 2 – 0.980 = 1.020

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

$m_{\text{CaSO}_{\text{4}}}=\text{1}\text{.020 mmol }\times \text{ 136}\text{.12 }\frac{\text{mg}}{\text{mmol}}=\text{1}\text{.38}\text{.8 mg}=\text{0}\text{.139 g}$

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.

The Solubilities of Salts of Weak Acids

In many chemical operations it is an advantage not only to be able to form a precipitate but to be able to redissolve it. Fortunately, there is a wide class of sparingly soluble salts which can almost always be redissolved by adding acid. These are precipitates in which the anion is basic; i.e., they are the salts of weak acids. An example of such a precipitate is calcium carbonate, whose solubility equilibrium is

CaCO₃(s) $\rightleftharpoons$ Ca²⁺(aq) + CO₃^2–(aq) (14.34)

If acid is now added to this solution, some of the carbonate ions become protonated and transformed into HCO₃^– ions. As a result, the concentration of the carbonate ion is reduced. In accord with Le Chatelier’s principle, the system will respond to this reduction by trying to produce more carbonate ions. Some solid CaCO₃ will dissolve, and the equilibrium will be shifted to the right. If enough acid is added, the carbonate-ion concentration in the solution can be reduced so as to make the ion product (Q = c_Ca2+ × c_CO32–) smaller than the solubility product K_sp so that the precipitate dissolves.

A similar behavior is shown by other precipitates involving basic anions. Virtually all the carbonates, sulfides, hydroxides, and phosphates which are sparingly soluble in water can be dissolved in acid. Thus, for instance, we can dissolve precipitates like ZnS, Mg(OH)₂, and Ca₂(PO₄)₃ because all the following equilibria

ZnS(s) $\rightleftharpoons$ Zn²⁺(aq) + S^2–(aq)

Mg(OH)₂(s) $\rightleftharpoons$ Mg²⁺(aq) + 2OH^–(aq)

Ca₂(PO₄)₃(s) $\rightleftharpoons$ 3Ca²⁺(aq) + 2PO₄^3–(aq)

can be shifted to the right by attacking the basic species S^2–, OH^–, and PO₄^3– with hydronium ions. Very occasionally we find an exception to this rule. Mercury(II) sulfide, HgS, is notorious for being insoluble. The solubility product for the equilibrium

HgS(s) $\rightleftharpoons$ Hg²⁺(aq) + S^2–(aq)

is so minute that not even concentrated acid will reduce the sulfide ion sufficiently to make Q smaller than K_sp.

Occasionally the shift in a solubility-product equilibrium caused by a decrease in pH may be undesirable. One example of this was mentioned in Sec. 12.6. Acid rainfall can occur when oxides of sulfur and other acidic air pollutants are removed from the atmosphere. In some parts of the United States pH values as low as 4.0 have been observed. These acid solutions dissolve marble and limestone (CaCO₃) causing considerable property damage. This is especially true in Europe, where some statues and other works of art have been almost completely destroyed over the last half century.

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