Illustrated Guide to Home Chemistry Experiments

Chapter 13 Laboratory: Chemical Equilibrium and Le Chatelier’s Principle 239

LABORATORY 1 3.3:

dETERmINE A SoLUBILITy pRodUCT CoNSTANT

The coexistence of an ionic solid and its

component ions in solution is one example of

a chemical equilibrium. This equilibrium can

be quantified using the solubility product

principle, which states that in a saturated

solution of an ionic compound, the product

of the molar activities, called the solubility

product constant or Ksp, has a constant value

at any particular temperature and pressure.

RIREEqU d EqUIpmENT ANd SUppLIES

£ goggles, gloves, and protective clothing

£ balance and weighing paper

£ beaker, 250 mL (2)

£ volumetric flask, 100 mL

£ eye dropper or disposable pipette

£ thermometer

£ stirring rod

£ funnel

£ filter paper

£ ring stand

£ support ring

£ wire gauze

£ burette clamp

£ gas burner

£ sodium chloride (~50 g)

£ potassium hydrogen tartrate (~2 g)

£ sodium hydroxide, 0.1000 m, standardized (~50 mL)

£ phenolphthalein indicator solution (a few drops)

Because ionic activities are very difficult to determine accurately
and molarities are easy to determine, solubility product
calculations are normally performed using molarity values. For
dilute solutions, using molarities rather than activities introduces
only tiny errors, because molarity and activity are nearly identical
in dilute solutions. For concentrated solutions, molarity and
activity may differ significantly, so solubility product calculations,
although useful for sparingly soluble ionic compounds, are of
limited use for very soluble ionic compounds.

Consider the sparingly soluble salt silver chloride. In a saturated
aqueous solution of silver chloride, the silver and chlorine are
present primarily as solvated silver ions and chlorine ions.
(Solvated ions are ions that have bound to solvent molecules.)
Undissociated molecular silver chloride exists in the solution,
but it is present in such a tiny amount that it can be ignored. The
following equilibrium equation describes the saturated silver
chloride solution:

AgCl(s) ⇔ Ag+(aq) + Cl–(aq)

The solubility product constant for this equilibrium is:

ksp = [Ag+] · [Cl–]

Square brackets in equilibrium expressions indicate molar
concentration, so this expression is a short way to say
that the solubility product constant is equal to the molar
concentration of the silver ions multiplied by the molar
concentration of the chloride ions. Because one molecule of
silver chloride dissociates into one silver ion and one chloride
ion, the concentrations of the silver ions and chloride ions are
identical. Assigning that concentration the value x simplifies the
expression to:

ksp = x · x = x^2

which means that the concentration of either ion is the

square root of the Ksp. That means that if we know the Ksp, we

can determine the molar concentration, and vice versa. For

example, if we know that the Ksp of silver chloride at a particular

temperature and pressure is 1.8 · 10–10, we can rewrite the

equilibrium expression as:

1.8 · 10–10 = x^2

Solving for x (taking the square root of the Ksp) tells us that the

molar concentrations of the silver ions and chloride ions are

both 1.3 · 10–5, so a saturated solution of silver chloride at this

temperature and pressure is 0.000013 M. The gram molecular

mass of silver chloride is 143.32 g/mol, so a saturated silver

chloride solution contains about 0.0019 g/L.