Solubility and precipitation

An insoluble salt in water is not inert: it is dissolving and crystallising at equal rates, and the balance point between them is an equilibrium constant like any other.

The previous two lessons applied the constant of the fourth lesson to proton transfer. This one applies it to a solid in contact with its saturated solution, which needs no new theory at all, only the rule that a pure solid has activity $1$ .

Saturation is an equilibrium

Drop silver chloride into water and a little dissolves, $AgCl(s) rightleftharpoons A(aq) + C(aq)$ . The solid is pure, so it does not appear in the constant, and what remains is the solubility product:

K_{s p} = [A] [C] = 1.77 \times 10^{- 10} at 25 °C

The equilibrium is dynamic, which can be shown directly. Add silver chloride labelled with radioactive silver-110 to a saturated solution of ordinary silver chloride and the radioactivity appears in the solution within minutes, although the amount of solid never changes. Nothing is at rest; the two rates are equal.

Two consequences follow from the form of the constant. The product of the ion concentrations is fixed whatever their individual values, so raising one must lower the other. And $K_{s p}$ depends on temperature alone, in the way the fifth lesson established, with $Δ_{sol} H$ setting the direction: most salts dissolve endothermically and so get more soluble when heated, while calcium sulfate and the other salts that scale a kettle are the exceptions, dissolving exothermically and precipitating as the water warms.

Molar solubility from the constant

The molar solubility $s$ is how many moles of the salt dissolve per cubic decimetre of saturated solution. Getting it from $K_{s p}$ means writing each ion concentration in terms of $s$ , remembering that a salt of formula $A_{x} B_{y}$ releases $x$ and $y$ ions per formula unit.

Example. Find the molar solubility of silver chloride, $K_{s p} = 1.77 \times 10^{- 10}$ , and of calcium fluoride, $K_{s p} = 3.9 \times 10^{- 11}$ , and express each in grams per litre. The molar masses are $143.32$ and $78.07$ g mol⁻¹.

For silver chloride, dissolving $s$ moles gives $[A] = [C] = s$ , so $K_{s p} = s^{2}$ and $s = \sqrt{1.77 \times 10^{- 10}} = 1.33 \times 10^{- 5}$ mol dm⁻³, or $1.9 \times 10^{- 3}$ g dm⁻³.

For calcium fluoride, dissolving $s$ moles gives $[C] = s$ and $[] = 2 s$ , so $K_{s p} = s (2 s)^{2} = 4 s^{3}$ . Then $s = (3.9 \times 10^{- 11} / 4)^{1 / 3} = 2.14 \times 10^{- 4}$ mol dm⁻³, or $0.0167$ g dm⁻³. Note the factor of two entering both as a coefficient and as a power; forgetting the power is the standard mistake here.

Now you. Silver chromate, $ACr$ , has $K_{s p} = 1.12 \times 10^{- 12}$ and a molar mass of $331.73$ g mol⁻¹. Find its molar solubility, and compare it with silver chloride's.

Answer

Two silver ions per formula unit, so $[A] = 2 s$ and $[Cr] = s$ , giving $K_{s p} = 4 s^{3}$ and $s = (1.12 \times 10^{- 12} / 4)^{1 / 3} = 6.5 \times 10^{- 5}$ mol dm⁻³, or $0.022$ g dm⁻³. Silver chromate has a solubility product a hundred and sixty times smaller than silver chloride's and is nearly five times more soluble.

That comparison is the point of the exercise. Solubility products can be compared directly only between salts of the same stoichiometry, because the exponent relating $s$ to $K_{s p}$ differs otherwise. Ranking a $1 : 1$ salt against a $2 : 1$ salt by their constants gives the wrong order, as it does here.

The common ion effect

Since $K_{s p}$ is a product, supplying one of the ions from elsewhere forces the other down, and the solid that has to precipitate to make that happen is the salt itself. This is the common ion effect, and it is Le Chatelier's principle in a form that can be computed exactly.

Example. Find the solubility of silver chloride in $0.010$ mol dm⁻³ sodium chloride, and compare with its solubility in pure water.

The dissolved silver chloride contributes a negligible amount of chloride compared with $0.010$ , so $[C] = 0.010$ and $[A] = K_{s p} / 0.010 = 1.77 \times 10^{- 8}$ mol dm⁻³. Since every dissolved silver ion means one dissolved formula unit, that is the solubility: $750$ times lower than the $1.33 \times 10^{- 5}$ in pure water. The approximation is safe because $1.77 \times 10^{- 8}$ is indeed tiny beside $0.010$ .

Now you. Find the solubility of calcium fluoride in $0.010$ mol dm⁻³ sodium fluoride, given $K_{s p} = 3.9 \times 10^{- 11}$ .

Answer

With $[] = 0.010$ fixed by the added salt, $[C] = K_{s p} / (0.010)^{2} = 3.9 \times 10^{- 11} / 1.0 \times 10^{- 4} = 3.9 \times 10^{- 7}$ mol dm⁻³, which is the solubility. That is $550$ times lower than in pure water. The suppression is stronger than for silver chloride at the same added concentration, because the fluoride concentration enters squared.

This is why a precipitate is washed with a dilute solution of one of its own ions rather than with pure water, and why gravimetric analysis adds an excess of the precipitating reagent: a hundredfold suppression of solubility is the difference between losing one per cent of the sample down the sink and losing none of it.

There is an opposite effect that the calculation above ignores. Adding an inert salt with no ion in common, such as potassium nitrate, actually raises the solubility slightly, because the extra ionic atmosphere lowers the activity coefficients of the dissolving ions, so a larger concentration is needed to reach the same activity product. It is a small correction in dilute solution and a real one in seawater.

Acid dissolves some salts and not others

The most useful prediction in this lesson comes from combining $K_{s p}$ with the previous two lessons. If the anion of a salt is a base, adding acid removes it from solution by protonating it, the ion product falls below $K_{s p}$ , and more solid dissolves. If the anion is not a base, acid does nothing at all.

Chloride is the conjugate base of a strong acid, so it is not a base in any useful sense, and silver chloride is as insoluble in dilute nitric acid as in water. Carbonate is the conjugate base of a weak acid with $p K_{a 2} = 10.33$ , so it is a strong base, and calcium carbonate dissolves readily in acid. The reaction runs to completion because the product escapes:

CaC(s) + 2 \to C + O + C(g)

Rainwater in equilibrium with atmospheric carbon dioxide is a weak acid at about pH $5.6$ , which is enough. Limestone caves, karst landscapes and the pitting of marble statues in polluted air are all this equation running for a long time, and the reverse of it, carbon dioxide escaping from groundwater as it drips into an air-filled cavity, is what builds a stalactite. Fluoride is a weaker base ( $p K_{a}$ of $HF$ is $3.17$ ), so calcium fluoride is intermediate: appreciably more soluble in strong acid, unaffected by weak.

Metal hydroxides are the extreme case, because their anion is hydroxide itself and the pH is the direct control. Magnesium hydroxide has $K_{s p} = 5.6 \times 10^{- 12}$ , so its saturated solution has $s = (5.6 \times 10^{- 12} / 4)^{1 / 3} = 1.12 \times 10^{- 4}$ mol dm⁻³ and $[O] = 2.24 \times 10^{- 4}$ , giving a pH of $10.35$ . That is milk of magnesia: alkaline enough to neutralise stomach acid, and self-limiting, because as acid is consumed the pH cannot rise past the saturated value.

Will it precipitate?

For a mixture that is not yet at equilibrium, the test is the one from the fourth lesson: compute the ion product $Q$ and compare it with $K_{s p}$ . If $Q > K_{s p}$ the solution is supersaturated and solid should form; if $Q < K_{s p}$ any solid present dissolves.

Where two salts could form, whichever needs the lower concentration of the added reagent precipitates first, and if the gap is wide enough the two can be separated almost completely. That is selective precipitation, the basis of classical qualitative analysis.

Example. A solution is $0.010$ mol dm⁻³ in both chloride and chromate. Silver nitrate is added slowly. Which precipitates first, and what fraction of the first ion remains when the second begins to come down? Use $K_{s p} = 1.77 \times 10^{- 10}$ for $AgCl$ and $1.12 \times 10^{- 12}$ for $ACr$ .

Silver chloride starts when $[A] = K_{s p} / [C] = 1.77 \times 10^{- 10} / 0.010 = 1.77 \times 10^{- 8}$ mol dm⁻³. Silver chromate starts when $[A] = \sqrt{K_{s p} / [Cr]} = \sqrt{1.12 \times 10^{- 10}} = 1.06 \times 10^{- 5}$ mol dm⁻³, six hundred times higher. So the chloride comes down first. At the moment the chromate begins, the remaining chloride is $1.77 \times 10^{- 10} / 1.06 \times 10^{- 5} = 1.7 \times 10^{- 5}$ mol dm⁻³, so $99.83$ per cent of the chloride has already precipitated. This is exactly the Mohr method for chloride, in which chromate is the indicator and the first permanent red tinge of silver chromate marks the end point.

Now you. Equal volumes of $0.0020$ mol dm⁻³ silver nitrate and $0.0020$ mol dm⁻³ sodium chloride are mixed. Does a precipitate form?

Answer

Mixing equal volumes halves both concentrations, so $[A] = [C] = 0.0010$ mol dm⁻³ and $Q = 1.0 \times 10^{- 6}$ . That exceeds $K_{s p} = 1.77 \times 10^{- 10}$ by nearly four orders of magnitude, so silver chloride precipitates until the product falls to $K_{s p}$ .

Whether it precipitates promptly is a separate question, and one for the kinetics half of this course. Solutions can stay supersaturated for a long time when there is no surface for a crystal to start on, which is why sodium acetate hand warmers hold litres of liquid far past saturation until a metal disc is clicked.

Complexing dissolves the insoluble

The other way to remove an ion from solution is to tie it up in a complex ion, which works on the cation as acid works on the anion. Silver ion binds two ammonia molecules with a formation constant $K_{f} = 1.7 \times 10^{7}$ , so the overall dissolution of silver chloride in ammonia is the sum of two equilibria and its constant is the product:

K = K_{s p} K_{f} = 1.77 \times 10^{- 10} \times 1.7 \times 10^{7} = 3.0 \times 10^{- 3}

In $1.0$ mol dm⁻³ ammonia that gives a solubility of about $0.049$ mol dm⁻³, nearly four thousand times the value in water, which is why ammonia clears a silver chloride precipitate and why photographic fixer, which uses thiosulfate with a much larger formation constant still, removes unexposed silver halide from film.

Some hydroxides do this to themselves. Aluminium hydroxide dissolves in acid as a base and in excess alkali as an acid, forming $[Al(OH]^{-}$ , so its solubility curve has a minimum near pH $6$ and rises in both directions. Water treatment exploits exactly that minimum.

A closing caution on all of these numbers. Solubility products are the equilibrium constants most affected by the activity coefficients set aside in the fourth lesson, since they involve multiply charged ions at appreciable ionic strength, and published values for the same salt can differ by a factor of two or three between sources. Treat a calculated solubility as good to an order of magnitude unless the ionic strength has been dealt with properly.

Moving the electron itself

Both families of reaction covered so far move a proton: from acid to base, or from acid to the anion of a salt. In every case the electrons stay where they were.

The last two lessons of this course are about reactions in which the electrons themselves change owner. They need their own bookkeeping, because an electron transferred between two atoms in the same molecule leaves no visible trace in the formula, and the accounting device that makes them visible is the next lesson.