Solution Equilibrium – The Solubility Product Constant (Text p. 482 – 493)
Check your understanding
1.
What is solubility?
2.
When is a substance soluable? When is it not?
3.
What is a saturated solution? Non-Saturated solution?
4.
What factors affect solubility?
5.
What is a molar solution?
6.
What is ionization?
Solution Equilibrium
As we learned in Acids and Bases, a strong acid or base is one that ionized 100%. In other words, all ionized and the equilibrium favours the formation of product.
Weak acids and bases do not ionize 100% and only partially, therefore, putting only a little hydrogen ion or hydroxide ion into solution.
Therefore, the equilibrium favours the formation of reactant.
For salts, when a solution is saturated , there is a point when the production of ions
(dissolving) is in equilibrium with the ions coming back together to form the salt.
This is called ____________________________ The more ions that form (ie. Very soluable), the greater the equilibrium constant (Keq is greater than 1) . The less ions that form (ie.
Low solubility), the lower the equilibrium constant (Keq is less than 1)
Example
Very soluable substance, dissociate 100%
Ex. NaCl Na+1 + Cl-1
(100 % ions until it is saturated)
Slightly soluable substance, does not dissociate to the same extent (ie. Less ions in solution that NaCl, at equilibrium)
Ex. AgCl Ag+1 + Cl-1
Notice the arrows, which is favoured at equilibrium . Some does dissolve, just a very small amount goes to ions.

K sp
or Solubility Product Constant
When a sparingly soluble ionic solid is dissolved in water to form a saturated solution the general equilibrium equation is as follows, where A is a positively charged ion and B is a negatively charged ion.
A a
B b ( s ) a A+ ( aq ) + b B¯ ( aq )
For very soluable substances, we know that 100% will form ions at equilibrium
At a given temperature, the equilibrium law for this reaction is given as:
Keq = [A+] a
[B-] b
[A a
B b
]
However, the term Keq[A a
B b
] can be replaced by a new constant, _____ , called the
_______________________.
Ksp = [A
+] a [B¯] b
The solubility product constant is the product of ion concentrations in a saturated solution. The solubility product constant takes into account the presence of the solid.
Ksp is the same as Keq and Ka / Kb, except it is the measure of the ion concentration at equilibrium (Saturation)

For highly soluable salts, Ksp will be greater than 1 o We don’t care about these salts, because for the most part, we can put a large amount into solution, and ions will form.
The slightly soluable salts are the ones we are concerned about

For slightly soluable salts, Ksp will be less than 1

Example o AgCl is a slightly soluable salt according to your rules. The Ksp value is
1.77 x 10 -10
Example 1
Write the dissociation equation and the expression for the solubility product constant for calcium hydroxide. (Ksp = 1.3 x 10-6)

Example 2
Write the solubility product expression for lead (II) phosphate (Pb3(PO4)2).
Example 3
What are the concentrations of ions for a saturated solution of calcium hydroxide?
(Ksp = 1.3 x 10-6)
Example 4
What is the Ksp value for barium sulfate, if at equilibrium, barium sulfate has a concentration of 3.87 x 10-5 M?
Example 5
If a saturated solution of magnesium fluoride has a Mg+2 ion concentration of
1.17 x 10 –3 M, what is the Ksp?

Solubility and solubility product are two different terms.

Solubility

Solubility product
Example 6
The solubility of PbF2 is 0.466 g/L. What is the value of the solubility product?
Example 7
The Ksp of magnesium hydroxide is 8.9 x 10
-12. What will be the equilibrium concentrations of the dissolved ions in a saturated 1 l solution of Mg(OH)2?
Example 8
What is the hydroxide concentration in a saturated 1 l solution of iron (III) hydroxide?
(Ksp = 4.00 x 10-38) Answer = 5.89 x 10-10 M
Example 9
What mass of silver (I) sulfate could be dissolved in 500 mL of water?
(Ksp = 1.20 x 10-5) Answer = 2.18 g

Solubility Rules and Precipitates
In previous chemistry courses you have learned the general rules for determining combinations of ions that have a low solubility. Insoluble compounds are generally described as those that precipitate upon mixing equal volumes of solutions which are 0.10 mol/L in the respective ions.
A precipitate is the
Example
Which of the following ions will precipitate with a solution of sodium chloride?
Copper (II) ion Lead (II) ion Silver ions Sulfide ions
Reaction Quotient and Ksp
Not all reactions produce a precipitate. If the volume of solution is large enough, and the amount of solute is small enough no precipitate will form. When a precipitate forms, the solution is saturated with the ions of the solute present. If the combined values of the precipitating ions are greater than the Ksp value, a precipitate will form.
We use the reaction quotient , Q , to predict whether a precipitate forms. The form of the reaction quotient is the same as the Ksp, just as it was with Keq and Ka/b values. By comparing the value of the reaction quotient with the solubility product, we can determine if an aqueous solution is saturated or unsaturated.
If,

Q = Ksp

Q>Ksp

Q<Ksp
We determine the Q value, from comparing our experimental concentrations to the Ksp value for the saturated solution

Example
The Ksp of lead (II) chloride is 1.6 x 10
-5. If 0.57 g of lead (II) chloride are added to
1500 mL of water, is the solution saturated? Assume no volume change.
Things to consider for determining precipitates:

The initial solutions must be soluable, or you do not have a solution

The chemical reaction is a double displacement reaction

The concentrations of the ions are what the stoichiometry states by the dissociation of the compound

When solutions are mixed, the volume changes, and so does the concentration of the ion

The net ionic equation will tell you what the Ksp expression should be

Find the Q and compare to the literature Ksp value
Example 1
If 20.0 mL of a 0.0010 mol/L silver nitrate solution is mixed with 20.0 mL of a 3.0 x
10-5 mol/L potassium bromide solution, does silver bromide (Ksp = 5.0 x 10
-13) precipitate? Assume the volumes are additive.

Example 2
100 ml of a 0.01 M potassium hydroxide solution is combined with 50 ml of a 0.1 M copper (II) nitrate solution. Will a precipitate form?
(Ksp for the precipitate = 1.60 x 10 –19)
Example 3
10 ml of a 0.001 M sodium chloride solution is added to 5 ml of 0.0001 M silver (I) nitrate solution. Will a precipitate form? (Ksp = 1.60 x 10
–10
) (4 marks)
Common Ion Effect
When an ionic compound dissolves in pure water, the initial concentration of each ion is zero. However, if an ionic compound dissolves in a solution that has an ion in common with the compound, this is not the case. Even though the starting concentrations may not be zero, the product of the ions must still equal the solubility product constant.
For example , how would the solubility of silver chloride in pure water change if we try dissolving it in tap water?

Le Chatelier's Principle predicts that the solubility of an ionic solid in a solution containing a common ion decreases its solubility. Let's see if this is supported by the calculations.
Example
Determine the solubility of silver chloride in pure water and in a solution of 0.10 mol/L sodium chloride.
The Ksp of AgCl is 1.7 x 10
-10.
Step 1
Find the solubility of the ions in pure water at saturation
Step 2
Set up an ICE Chart for the addition of the NaCl solution. What is the common ion? What is its concentration?
Step 3
Substitute the values from your ICE chart into your Ksp equation.
Step 4
Solve for x. Since x is small, disregard as it will have little impact on the final answer.
Step 5
Find the value of solubility of the NaCl, using the chloride ion value

Example 2
The Ksp of lead (II) chloride, PbCl2, is 1.6 x 10
-5. What is the solubility of lead (II) chloride in a 0.10 mol/L solution of magnesium chloride, MgCl2?
MgCl2 is a highly soluable salt, with 100% ionization. Therefore, the solubility of PbCl2 will be less, due to the common ion in solution.

Solutions Review
1.
Define solubility.
2.
What happens to the solubility of solids when the temperature increases? Explain why.
3.
What happens to the solubility of gases when the temperature decreases? Explain why.
4.
A solution of sodium carbonate contains 53.00 g of solute in 215 ml of water.
What is the molarity of the solution?
5.
What are the concentrations of the ions in solution for a saturated solution of silver (I) chloride? (Ksp = 1.60 x 10
–10
)
6.
The Ksp of magnesium carbonate at 25 o
C is 2.00 x 10
–8
. What mass of magnesium carbonate could be dissolved in 1L of water?
7.
If the [Cu
+2
] in copper (I) oxide is 2.00 x 10
–5
M at equilibrium, what is the Ksp of Copper (I) oxide?
8.
How many grams of cadmium (II) carbonate can you dissolve in 1 L of water?
(Ksp=5.20 x 10
–12
)
9.
100 ml of a 0.01 M potassium hydroxide solution is combined with 50 ml of a 0.1
M copper (II) nitrate solution. Will a precipitate form?
(Ksp for the precipitate = 1.60 x 10
–19
)
10.
What is the maximum amount of [Sr +2 ] that can be dissolved in a 0.02 M solution of potassium sulfate, without precipitating strontium sulfate?
(Ksp SrSO
4
= 3.20 x 10
–7
)
11.
10 ml of a 0.001 M sodium hydroxide solution is added to 5 ml of 0.0001 M calcium nitrate solution. Will a precipitate form? (Ksp = 5.02 x 10
–6
) (4 marks)
12.
You need to lower the concentrations of lead in your water supply. If the common ion in our water supply is the chloride ion, what could you add to the water supply to remove lead? Explain your reasoning. (2 marks)
13.
We use fluoride in our water to strengthen the enamel on our teeth. What is the maximum concentration of [F
-
] you could add to our water to ensure that you would not get a precipitate with the lead, calcium and magnesium we have in our water? (2 marks) a.
Ksp for lead (II) fluoride = 4.00 x 10
-8 b.
Ksp for calcium fluoride = 4.00 x 10
-11 c.
Ksp for magnesium fluoride = 6.40 x 10
-9