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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¯] = s. Substituting into the Ksp expression:
1.2 x 10¯11 = (1.1 x 10¯10) (s)2x = 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) (s)2x = 5.91657 x 10-8 M (I kept a few guard digits.)
3) Compute the pH:
pOH = 7.228pH = 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 = (s)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¯]21.2 x 10¯11 = (s) (6.10 x 10¯5)2
s = 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¯]36.7 x 10-31 = (s) (3s)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/LAll 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)39.17 x 10-7 g/L divided by 103 g/mol = 8.90 x 10-9 MNaOH
4.0 x 10-4 g / 120 L = 3.33 x 10-6 g/L3.33 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)3x = 5.15 x 10-30
Cr(OH)3 precipitates.
Q is a 'reaction quotient.' You might have to look that up.
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 = (s) (1.0 x 10¯3)3x = 6.7 x 10-22 M
Problem #6: The Ksp of AgBr is 5.4 x 10¯13 at 25 °C. Calculate the molar solubility of AgBr in 0.050 M AgNO3(aq) at 25 °C.
Solution:
Ksp = [Ag+] [Br¯]The solution already contains 0.050 M Ag+ from the dissociation of the AgNO3.
The solution will also contain additional Ag+, due to the dissociation of the AgBr. This value is small compared to 0.050 M, it can be ignored.
Substitute into Ksp equation:
5.4 x 10¯13 = (0.050) (s)
s = (5.4 x 10¯13) / 0.050
s = 1.08 x 10¯11 M
To two significant digits, the bromide concentration is 1.1 x 10¯11 mol/L
If you decided that the additional amount of silver ion CANNOT be ignored, you would have this:
5.4 x 10¯13 = (0.050 + s) (s)
The produces a quadratic:
s2 + 0.05s - 5.4 x 10¯13 = 0
Substituting into a quadratic solver gives:
1.0799999997667 x 10¯11
In other words, the ignoring was a perfectly valid thing to do.