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.

Present in silver chloride are silver ions (Ag+) and chloride ions (Cl¯). Silver nitrate (which is soluble) has silver ion in common with silver chloride. Sodium chloride (also soluble) has chloride ion in common with silver chloride.

In fact, mixing sufficiently concentrated solutions of AgNO3 and NaCl will produce a precipitate of AgCl. In order to be sufficiently concentrated, the product of the [Ag+] and the [Cl¯] must exceed the Ksp of 1.77 x 10¯10.


Example #1: AgCl will be dissolved into a solution with 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. So, on to the solution . . .

The dissociation equation for AgCl is:

AgCl (s) <===> Ag+ (aq) + Cl¯ (aq)

The Ksp expression is:

Ksp = [Ag+] [Cl¯]

This is the equation we must solve. First we put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]

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 'x' amount from the dissolving AgCl. Of course, [Ag+] is 'x.'

Substituting, we get:

1.77 x 10¯10 = (x) (0.0100 + x)

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

1.77 x 10¯10 = (x) (0.0100)

and

x = 1.77 x 10¯8 M

There is another reason why neglecting the 'x' in '0.0100 + x' 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 'x' and '0.0100 + x' 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: The Ksp for silver carbonate is 8.4 x 10¯12. The concentration of carbonate ions in a saturated solution is 1.28 x 10¯4 M. What is the concentration of silver ions?

Solution:

Dissociation equation:

Ag2CO3 (s) <===> 2 Ag+ (aq) + CO32¯ (aq)

Ksp expression:

Ksp = [Ag+]2 [CO32¯]

Let us substitue into the Ksp expression:

8.4 x 10¯12 = (x)2 (1.28 x 10¯4)

Note: I could have used (1.28 x 10¯4 + 0.5x) for the carbonate. See above discussion for why the 0.5x can be dropped.

Divide both sides by 1.28 x 10¯4 and then take the square root:

[Ag+] = x = 2.56 x 10¯4 M

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?

Solution:

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

Dissociation equation:

M(OH)2 (s) <===> M2+ (aq) + 2 OH¯ (aq)

Ksp expression:

Ksp = [M2+] [OH¯]2

Let us substitue into the Ksp expression:

x = (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.

The answer:

Ksp = 1.0 x 10¯7

Example #4: What is the solubility of AgI in a 0.274-molar solution of NaI. (Ksp of AgI = 8.52 x 10¯17)

Solution:

Dissociation equation:

AgI (s) <===> Ag+ (aq) + I¯ (aq)

Ksp expression:

Ksp = [Ag+] [I¯]

Let us substitue into the Ksp expression:

8.52 x 10¯17 = (x) (0.274)

The answer:

[Ag+] = 3.11 x 10¯16 M

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

Solution:

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

Dissociation equation:

Ca(OH)2 (s) <===> Ca2+ (aq) + 2OH¯ (aq)

Ksp expression:

Ksp = [Ca2+] [OH¯]2

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

4.68 x 10¯6 = (x) (0.172 + 2x)2

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

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: The solubility product of Mg(OH)2 is 1.2 x 10¯11. What minimum OH¯ concentration must be attained (for example, by adding NaOH) to decrease the Mg2+ concentration in a solution of Mg(NO3)2 to less than 1.1 x 10¯10 M?

Solution:

Ksp expression:

Ksp = [Mg2+] [OH¯]2

We set [Mg2+] = 1.1 x 10¯10 and [OH¯] = x. Substituting into the Ksp expression:

1.2 x 10¯11 = (1.1 x 10¯10) (x)2 x = 0.33 M

Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg2+] to less than 1.1 x 10¯10 M.


Example #7: Calculate the pH at which zinc hydroxide just starts to precipitate from a 0.00857 M solution of zinc nitrate. Ksp for zinc hydroxide = 3.0 x 10-17

Solution:

1) Ksp expression:

Ksp = [Zn2+] [OH¯]2

2) Substitute and solve for [OH¯]:

3.0 x 10-17 = (0.00857) (x)2

x = 5.91657 x 10-8 M (I kept a few guard digits.)

3) Compute the pH:

pOH = 7.228

pH = 6.772

Note how zinc hydroxide would precipitate even when the solution is slightly acidic.


Example #8: What is the solubility, in moles per liter, of AgCl (Ksp = 1.77 x 10-10) in 0.0300 M CaCl2 solution?

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 = (x) (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 #9: Calculate the number of moles of Ag2CrO4 that will dissolve in 1.00 L of 0.010 M K2CrO4 solution. Ksp for Ag2CrO4 = 9.0 x 10-12.

Solution:

1) Concentration of dichromate ion from potassium chromate:

0.010 M

2) Calculate solubility of Ag+:

Ksp = [Ag+]2 [CrO42¯]

9.0 x 10-12 = (x)2 (0.010)

x = 3.0 x 10-5 M

Since there is a 2:1 ratio between the moles of aqueous silver ion and the moles of silver chromate that dissolved, 1.5 x 10-5 M is the molar solubility of Ag2CrO4 in 0.010 M K2CrO4 solution.

Since we were asked for the moles of silver chromate that would disolve in 1.00 L, the final answer is:

1.5 x 10-5 mol

Example #10: What is the maximum concentration of Mg2+ ion that can remain dissolved in a solution that contains 0.7147 M NH3 and 0.2073 M NH4Cl? (Ksp for Mg(OH)2 is 1.2 x 10¯11; Kb for NH3 is 1.77 x 10¯5)

Solution:

1) Use the acid base data supplied to calculate [OH¯]:

Kb = ([NH4+] [OH¯]) / [NH3]

1.77 x 10¯5 = [(0.2073) (x)] / 0.7147

x = 6.10 x 10¯5 M

2) Use the Ksp expression to calculate the [Mg2+]:

Ksp = [Mg2+] [OH-]2

1.2 x 10¯11 = (x) (6.10 x 10¯5)2

x = 3.2 x 10¯3 M


Example #11: 1.1 x 10-4 g of Cr(OH)3 is added to 120 L of water at 25 °C. Will it all dissolve? (Ksp = 6.7 x 10-31)

Solution:

1) Solve Ksp expression for the molar solubility:

Ksp = [Cr3+] [OH¯]3

6.7 x 10-31 = (x) (3x)3

x = 1.255 x 10-8 M

2) Convert to grams per liter:

1.255 x 10-8 mol/L times 103.0 g/mol = 1.29 x 10-6 g/L

3) Check the problem's data:

1.1 x 10-4 g / 120 L = 9.17 x 10-7 g/L

All the Cr(OH)3 dissolves.

Example #11 - Part 2: 4.0 x 10-4 g of NaOH is added. Will a precipitate form?

Solution:

1) Convert g/L to mol/L:

Cr(OH)3
9.17 x 10-7 g/L divided by 103 g/mol = 8.90 x 10-9 M

NaOH

4.0 x 10-4 g / 120 L = 3.33 x 10-6 g/L

3.33 x 10-6 g/L divided by 40.0 g/mol = 8.33 x 10-8 M

2) Calculate a Qsp:

x = (8.90 x 10-9) (8.33 x 10-8)3

x = 5.15 x 10-30

Cr(OH)3 precipitates.

Example #11 - Part 3: Calculate the molar solubility of Cr(OH)3 in a solution buffered at pH = 11.00

Solution:

1) Determine the [OH¯]:

pOH = 14.00 - 11.00 = 3.00

[OH¯] = 10¯pOH = 10¯3.00 = 1.0 x 10¯3 M

2) Determine the molar solubility:

6.7 x 10-31 = (x) (1.0 x 10¯3)3

x = 6.7 x 10-22 M


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