The Common Ion Effect

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The solubility of insoluble substances can be decreased by the presence of a common ion. AgCl will be our example.

AgCl is an ionic substance and, when a tiny bit of it dissolves in solution, it dissociates 100%, into silver ions (Ag+) and chloride ions (Cl¯).

Now, consider silver nitrate (AgNO3). When it dissolves, it dissociates into silver ion and nitrate ion. In the chemistry world, we say that silver nitrate has silver ion in common with silver chloride. Now, consider sodium chloride. It produces sodium ion and chloride ion in solution and we say NaCl has chloride ion in common with silver chloride.

What we do is try to dissolve a tiny bit of AgCl in a solution which ALREADY has some silver ion or some chloride ion (never both at the same time) dissolved in it. What will happen is that the solubility of the AgCl is lowered when compared to how much AgCl dissolves in pure water.

We call this the common ion effect.

Example #1: AgCl will be dissolved into a solution which is ALREADY 0.0100 M in chloride ion. What is the solubility of AgCl?

By the way, the source of the chloride is unimportant (at this level). Let us assume the chloride came from some dissolved sodium chloride, sufficient to make the solution 0.0100 M.


1) The dissociation equation for AgCl is:

AgCl (s) ⇌ Ag+ (aq) + Cl¯ (aq)

2) The Ksp expression is:

Ksp = [Ag+] [Cl¯]

3) The above is the equation we must solve. First we put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]

4) Now, we have to reason out the values of the two guys on the right. The problem specifies that [Cl¯] is already 0.0100. I get another 's' amount from the dissolving AgCl. By the 1:1 stochiometry between silver ion and chloride ion, the [Ag+] is 's.' Substituting, we get:

1.77 x 10¯10 = (s) (0.0100 + s)

5) This will wind up to be a quadratic equation which is solvable via the quadratic formula. However, there is a simplified way to solve this problem. We reason that 's' is a small number, such that '0.0100 + s' is almost exactly equal to 0.0100. If we were to use 0.0100 rather than '0.0100 + s,' we would get essentially the same answer and do so much faster. So the problem becomes:

1.77 x 10¯10 = (x) (0.0100)


x = 1.77 x 10¯8 M

There is another reason why neglecting the 's' in '0.0100 + s' is OK. It turns out that measuring Ksp values are fairly difficult to do and, hence, have a fair amount of error already built into the value. So the very slight difference between 's' and '0.0100 + s' really has no bearing on the accuracy of the final answer. Why not? Because the Ksp already has significant error in it to begin with. Our "adding" a bit more error is insignificant compared to the error already there.

Example #2: What is the solubility of AgI in a 0.274-molar solution of NaI. (Ksp of AgI = 8.52 x 10¯17)


1) Dissociation equation:

AgI (s) ⇌ Ag+ (aq) + I¯ (aq)

2) Ksp expression:

Ksp = [Ag+] [I¯]

3) Let us substitue into the Ksp expression:

8.52 x 10¯17 = (s) (0.274 + s)

4) The answer (after neglecting the +s in 0.274 + s:

[Ag+] = 3.11 x 10¯16 M

By the 1:1 stoichiometry between silver ion and AgI, the solubility of AgI in the solution is 3.11 x 10¯16 M

5) By the way, the solubility of AgI in pure water is this:

8.52 x 10¯17 = (s) (s)

x = 9.23 x 10-9 M

The solubility of the AgI has been depressed by a factor of a bit less than 30 million times.

Example #3: The molar solubility of a generic substance, M(OH)2 in 0.10 M KOH solution is 1.0 x 10¯5 mol/L. What is the Ksp for M(OH)2?


In this case, we are being asked for the Ksp, so that is where our unknown will be. That means the right-hand side of the Ksp expression (where the concentrations are) cannot have an unknown.

1) Dissociation equation:

M(OH)2 (s) ⇌ M2+ (aq) + 2OH¯ (aq)

2) Ksp expression:

Ksp = [M2+] [OH¯]2

3) Let us substitue into the Ksp expression:

Ksp = (1.0 x 10¯5) (0.10)2

The 1.0 x 10¯5 comes from the molar solubility information, coupled with the fact that for every one M(OH)2, one M2+ is produced.

Also, we could have used (0.10 + 2.0 x 10¯5) M for the [OH¯]. However, the 2.0 x 10¯5 M, being much smaller than 0.10, is generally ignored.

4) The answer:

Ksp = 1.0 x 10¯7

Example #4: What is the solubility, in moles per liter, of AgCl (Ksp = 1.77 x 10-10) in 0.0300 M CaCl2 solution?


1) Concentration of chloride ion from calcium chloride:

0.0300 M x 2 = 0.0600 M

from here:

CaCl2(s) ---> Ca2+(aq) + 2Cl¯(aq)

2) Calculate solubility of Ag+:

Ksp = [Ag+] [Cl¯]

1.77 x 10-10 = (s) (0.0600)

x = 2.95 x 10-9 M

Since there is a 1:1 ratio between the moles of aqueous silver ion and the moles of silver chloride that dissolved, 2.95 x 10-9 M is the molar solubility of AgCl in 0.0300 M CaCl2 solution.

Example #5: What is the solubility of Ca(OH)2 in 0.0860 M Ba(OH)2?


1) Dissociation equation:

Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH¯(aq)

2) Ksp expression:

Ksp = [Ca2+] [OH¯]2

3) The Ksp for Ca(OH)2 is known to be 4.68 x 10¯6. We set [Ca2+] = s and [OH¯] = (0.172 + 2s). Substituting into the Ksp expression:

4.68 x 10¯6 = (s) (0.172 + 2s)2

By the way, Ba(OH)2 is a strong base so [OH¯] = 2 times 0.0860 = 0.172 M

Ignoring the "2s," we find s = 1.58 x 10¯4 M

Since there is a 1:1 molar ratio between calcium ion and calcium hydroxide, 1.58 x 10¯4 M is the concentration of the calcium hydroxide.

Comment: There are several different values floating about the Internet for the Ksp of Ca(OH)2. I got mine from the CRC Handbook, 73rd Edition, pg. 8-43.

Example #6: How many grams of Fe(OH)2 (Ksp = 1.8 x 10¯15) will dissolve in one liter of water buffered at pH = 12.00?


1) The chemical equation:

Fe(OH)2 ⇌ Fe2+ + 2OH¯

2) The Ksp expression:

Ksp = [Fe2+] [OH¯]2

3) pH of 12.00 means pOH of 4.00. Which means this:

[OH¯] = 1.0 x 10¯4 M

4) The word buffer means that, for all intents and purposes, the [OH¯] will remain constant as some Fe(OH)2 dissolves.

5) Solve the Ksp:

1.8 x 10¯15 = [Fe2+] [1.0 x 10¯4]2

[Fe2+] = 1.8 x 10¯7 M

6) The Fe(OH)2 that dissolves is in a 1:1 molar ratio with the Fe^2+, so we see that 1.8 x 10¯7 mol of Fe(OH)2 dissolves in our 1.00 L of solution.

(1.8 x 10¯7 mol) (89.8588 g/mol) = 0.000016174584 g

1.6 x 10¯5 g (rounded to two sig figs)

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