The Common Ion Effect
Problems 1 - 10

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Problem #1: 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.


Problem #2: 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.


Problem #3: 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

Problem #4: 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


Problem #5: 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.

Problem #5 - 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.

Problem #5 - 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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