Solubility Product Constants

To round out Lesson 4, we are going to return to a topic from Lesson 3: the solubility of ionic compounds. Early in Lesson 3, you learned that ionic compounds are either soluble or insoluble in water, and that you can apply solubility rules to determine which category a compound falls into. However, at that point we alluded to the fact that this is an oversimplification: even "insoluble" compounds will generally dissolve into water at least a tiny bit, and even "soluble" ionic compounds eventually reach a solubility limit where no more solute can be dissolved into a solution.

We are now going to explore the nature of solubility for ionic compounds in a more rigorous way. First, we will learn more about the process of dissolution, then we will learn some mathematical techniques related to ionic compound solubility.

Dynamic Equilibrium

If there is enough solid present, the solution will eventually become saturated, reaching its solubility limit. At this point, both these processes - ions separating from the solid, and ions redepositing onto it - continue happening, with equal speed in both directions. The effect is that the amount of ions present in solution (and the amount present in the solid) remains constant, but any individual ion may continue shuttling back and forth between the two phases.

This is a situation known as dynamic equilibrium. The term "equilibrium" refers to the fact that the concentration is constant over time, while the term "dynamic" refers to the fact that individual ions are still changing their phase, even though the overall amount of ions in each phase stays constant. This is the first example of dynamic equilibrium we are encountering, but it will not be the last. Dynamic equilibrium in one form or another will be important in all our remaining lessons.

To represent the dynamic equilibrium of a saturated solution, we can use "double arrows" or "equilibrium arrows." These indicate the fact that, for any individual ion, the process of dissolution goes in both directions, and that each direction happens with equal speed overall.

NaCl (s) ⇌ Na⁺ (aq) + Cl^- (aq)

This form of equilibrium has some interesting effects we can observe. For example, the solubility of NaCl is about 6 M, so if we take a beaker of water and dump in NaCl until the solution is saturated and solid NaCl sits on the bottom of the beaker, the concentration in the solution is 6 M. Let's say we then take a different source of chloride ions, like HCl, and add it to the saturated solution, enough to make the the Cl^- concentration 8 M.

If we do this, what we will observe is that some salt will crystallize, falling out of solution as solid. And instead of having a sodium ion concentration of 6 M and a chloride concentration of 8 M, if we measure the concentrations of ions, we would find the sodium ion to have a concentration of about 5 M and the chloride a concentration of about 7 M. Why does this happen?

Well, the rate at which ions dissolve from the solid (the "forward reaction") does not change. If I add a bunch of Cl^- to my solution, though, then collisions with the solid become more frequent, so the "reverse reaction" gets faster. So solid is being formed faster than it is disappearing ... at least until the sodium and chloride concentrations drop enough for things to balance back out.

Solubility Products

As it turns out, the solubility of an ionic compound like NaCl is controlled by a value: the solubility product constant (or K_sp). For NaCl, K_sp is equal to a value of 36, and it is equal to the product of the ion concentrations in the solution.

K_sp = [Na⁺]*[Cl^-]

As long as there is only sodium chloride in solution, the concentrations are equal, so the solubility is the square root of 36, or 6 M. However, when we added enough chloride to bring the concentration up to 8 M, then the math no longer checked out; 6*8 is 48, not 36. So some NaCl crystallized until the concentrations were down to the correct levels. The exact values end up being [Na⁺] = 5.08 M and [Cl^-] = 7.08 M. Confirm for yourself that these values multiply to give 36.

Likewise, we can calculate what the concentration of sodium ions would be if the chloride concentration were, say, 9.0 M.

36 = [Na⁺]*9.0 M

[Na⁺] = 4.0 M

The same line of reasoning can be used with any salt that dissolves in water, even if it dissolves only a very small amount. For example, silver chloride will dissolve somewhat in water. However, it reaches saturation very quickly -that is, when the concentrations of silver and chloride ions are about 1.3*10^-5 M. As with NaCl, we can write an equation for the dissociation of silver chloride with equilibrium arrows, and we can write an expression for its K_sp value.

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq)

K_sp = [Ag⁺]*[Cl^-]

Now you can get some more practice with calculations: if the solubility of AgCl in water is 1.3*10^-5 M, then what is its K_sp equal to? All we have to do is take the 1.3*10^-5 M and plug it in for both the concentration of silver and the concentration of chloride (and we ignore the units, so just drop the "M").

K_sp = 1.3*10^-5 * 1.3*10^-5

K_sp = 1.69*10^-10

As before, we can use the expression for K_sp to solve for the concentration of one ion, given the concentration of the other. So, for example, if we added enough chloride to the solution to make the concentration 0.500 M, what would the new silver concentration become?

1.69*10^-10 = [Ag⁺]*0.500 M

[Ag⁺] = 3.38*10^-10

When the formula for a salt contains more than just one of each ion, the solubility product equation gets a little more complicated. Let's use PbCl₂ as an example. When PbCl₂ dissolved in water, two ions of Cl^- are released for every one ion of Pb²⁺. In a saturated solution, the dissolution is reversible, so we write our equation as shown below.

PbCl₂(s) ⇌ Pb²⁺ (aq) + 2 Cl^- (aq)

The coefficient of "2" on the chloride ion changes our expression for K_sp just a little; the 2 becomes an exponent, meaning the chloride concentration gets squared in the expression.

K_sp = [Pb²⁺]*[Cl^-]²

The reason for this becomes clear if you write the first equation a little differently.

PbCl₂(s) ⇌ Pb²⁺ (aq) + Cl^- (aq) + Cl^- (aq)

Written this way, if we follow our original pattern of multiplying everything together, we get

K_sp = [Pb²⁺]*[Cl^-]*[Cl^-]

which simplifies to

K_sp = [Pb²⁺]*[Cl^-]²

Example Problems

1. For the reaction above, if the value of K_sp is 2.01*10^-5 and the concentration of chloride ions is 2.5*10^-5 M, what is the concentration of lead (II) ions?

2. Write a dissolution equation and K_sp expression for Fe(OH)₃

3. If the value of K_sp for Fe(OH)₃ is 2.8*10^-39 calculate the concentration of iron (III) if the hydroxide concentration is 7.2*10^-7 M

Example Problem Answers

1. For the reaction above, if the value of K_sp is 2.01*10^-5 and the concentration of chloride ions is 0.025 M, what is the concentration of lead (II) ions?

2.01*10^-5 = [Pb²⁺]*(0.025)²

[Pb²⁺] = 2.01*10^-5/(0.025)²

[Pb²⁺] = 0.032 M

2. Write a dissolution equation and K_sp expression for Fe(OH)₃

Fe(OH)₃ (s) ⇌ Fe³⁺ (aq) + 3 OH^- (aq)

K_sp = [Fe³⁺]*[OH^-]³

3. If the value of K_sp for Fe(OH)₃ is 2.8*10^-39 calculate the concentration of iron (III) if the hydroxide concentration is 7.2*10^-7 M

2.8*10^-39 = [Fe³⁺]*(7.2*10^-7)³

[Fe³⁺] = 2.8*10^-39/(7.2*10^-7)³

[Fe³⁺] = 7.5*10^-21 M