CHEM 115 HOMEWORK 1 NAME_________________
Due Wed. ApriL 11, 2001 COMP. HMWRK: Solubility Eq.
1-3Or MW citric acid = 192.124
A 0.688 m solution of citric acid (H3C6H5O7) in water has a density of 1.049g/cm3.
Calculate the
1.37 mol/L x 192.124 g/mol = 263.21 g/L
(263.21 g citric acid/L / 1100 g/L solution) x 100% = 23.93%
mass % = |_23.9%_______
b. molality / Molarity = moles solute / kg solvent
g H2O = 1100 g solution – 263.2 g citric acid= 836.8 g = 0.8368 kg
1.37 mol / 0.8368 kg = 1.637 m
m = |__1.64 m___________
c. mole fraction citric acid = mol solute / total moles
1.37 mol / (1.37 + 835.8/18.02) = 1.37 / (1.37 + 46.44) = 1.37 mol / 47.807 mol = 0.02896
X = |_0.0287____________
0.688 mol x 263.21 g/mol = 132.18 g citric acid per 1000 g H2O
100% x 132.2 g citric acid / (1000 g H2O + 132.2 g citric) = 100% x 132.2/1132.2 = 11.67%
mass% = |_11.7%________
D = M/V; V = M/D = 1132.2 g solution / 1.049 g/mL = 1079.3 mL = 1.0793 L
0.688 mol / 1.079 L = 0.6374 M
M = |__0.637 M_________
0.688 mol / (0.688 mol + 1000 g H2O/18.02 g/mol) = 0.688 / )0.688+55.49) = 0.688 / 56.18 = 0.01224
X = |__0.0122 __________
CH3NH2 = methylamine = B; CH3NH3+ = BH+; Cl
- is a spectator ionConjugate weak base-acid pair à buffer; B + H2O = BH+ + OH-
Kb = 4.4 x 10-4 = [BH+][OH-] / [B] = [0.70 + x][x] / [0.50-x] ~ [0.70][x] / [0.50]
x ~ 4.4 x 10-4 x 0.50 / 0.70 ~ 3.143 x 10-4 ~ [OH-]; approx okay, 3.1 x 10-4 <<0.50
pOH = 3.503; pH = 10.497
pH = |__10.50___________
|
NaOH |
B |
BH+ |
|
|
mmol |
10.0 ml x 1.0 M = 10 |
100 ml x 0.50 M = 50 |
100 mL x 0.70 M = 70 |
|
BH+ + |
OH- à |
B + |
H2O |
|
70 – 10 = 60 mmol |
10 - 10 = 0 mmol (LR) |
50 + 10 = 60 mmol |
Kb = 4.4 x 10-4 = [BH+][OH-] / [B] = [60mmol/110mL + x][x] / [60mmol/110mL-x] ~ [60][x] / [60]
x ~ 4.4 x 10-4 ~ [OH-] << 60/110 = 0.545; approx ok; pOH = 3.357; pH = 10.64
pH= |__10.64_________
4. Calculate the pH of a solution after 10.0 ml of 1.0 M HCl are added to the buffer in problem # 2
Assume 100 mL of buffer again. Add acid to buffer; should become slightly more acidic
|
HCl |
B |
BH+ |
|
|
mmol |
10.0 ml x 1.0 M = 10 |
100 ml x 0.50 M = 50 |
100 mL x 0.70 M = 70 |
|
B + |
H+ à |
BH+ |
|
50 – 10 = 40 mmol |
10 – 10 = 0 mmol (LR) |
70 + 10 = 80 mmol |
Kb = 4.4 x 10-4 = [BH+][OH-] / [B] = [80mmol/110mL + x][x] / [40mmol/110mL-x] ~ [80][x] / [40]
x ~ 4.4 x 10-4 x 40/80 ~ 2.2 x 10-4~ [OH-] ; approx ok; pOH = 3.658; pH = 10.34
pH = |__10.34___________
5. Calculate the pH of a 1.50 M NH4ClO4 solution. à NH4+ + ClO4
-; ClO4 - is conj to strong acidbut NH4+ is conj acid to weak base; thus hydrolysis. Solution should be acidic
|
NH4+ + |
H2O à |
H3O+ + |
NH3 |
|
1.50 - x |
x |
x |
Ka = Kw / Kb = 1.0 x 10-14 / 1.8 x 10-5 = 5.56 x 10-10 = [H3O+][NH3] / [NH4+] = [x][x] / [1.50 –x]
~ x2 / 1.50
x ~ (5.56 x 10-10 * 1.50)1/2 ~ (8.34 x 10-10)1/2 ~ 2.88 X 10-5 ~ [H3O+]; approx ok; pH ~ 4.539
pH = |__4.54___________
6. Calculate the molar solubility of Ba(IO3)2 if its solubility product is 1.57 x 10-9.
|
Ba(IO3)2 (s) = |
Ba2+ + |
2 IO3- |
|
(-x) |
x |
2x |
Ksp = [Ba2+][IO3-]2 = [x][2x]2 = 4x3 = 1.57 x 10-9
X = (1.57 x 10-9 / 4)1/3 = (3.925 x 10-10)1/3 = 7.322 x 10-4
Sol. = |_7.32 x 10-4 M_____
7.a. Calculate the solubility product Ksp for BaSO4 if the solubility is 0.0245 g of BaSO4 in 100 ml of water at 25 oC. MW = 233.336; 100 mL = 0.100 L. Ksp must be in M concentrations
(0.0245 g / 233.336 g/mol) /0.100L = 1.050 x 10-3M
|
BaSO4 (s) = |
Ba2+ + |
SO42- |
|
(-1.05 x 10-3) |
1.05 x 10-3 |
1.05 x 10-3 |
Ksp = [Ba2+][SO42-] = [1.05 x 10-3 ][1.05 x 10-3] = 1.10 x 10-6
Ksp. = |__1.10 x 10-6 ____
This is a case such as described in the book such that you would expect ion pairing (“high” charges) and this Ksp to be significantly different than that you find in a table of thermodynamic Ksp’s.
b. Calculate the solubility product for Co(IO3)2 if its molar solubility is 1.1 x 10-2 mol/L at 18oC.
|
Co(IO3)2 (s) = |
Co2+ + |
2 IO3- |
|
(- 1.1 x 10-2) |
1.1 x 10-2 |
2(1.1 x 10-2) |
Ksp = [Co2+][IO3-]2 = [1.1 x 10-2][2 x 1.1 x 10-2)2 = 4(1.1 x 10-2)3 = 5.32 x 10-6
Sol. Prod. = |_5.3 x 10-6______
8. Which salt is more soluble, AgCN or Zn(CN)2? If different type of chemical formula, you can’t
AgCN, Ksp = 2.2 x 10-16 Zn(CN)2, Ksp = 3 x 10-16 compare Ksp’s directly.
|
AgCN(s) = |
Ag+ + |
CN- |
Zn(CN)2 (s) = |
Zn2+ + |
2 CN- |
|
|
(-x) |
x |
x |
(-x) |
x |
2x |
Ksp = 2.2 x 10-16 = [Ag+][CN-] = [x][x]; x = (2.2 x 10-16)1/2 = 1.48 x 10-8M
Ksp = 3 x 10-16 = [Zn2+][CN-]2 = [x][2x]2 = 4 x3; x = (3 x 10-16 / 4)1/3 = 4.2 x 10-6M, which is larger so that Zn(CN)2 is more soluble.
9. Chalk is CaCO3 and at 25oC its Ksp = 4.5 x 10
–9, What is the molar solubility of calcium carbonate? How many grams will dissolve in 100 ml of aqueous solution? MW = 100.09 g/mol|
CaCO3(s) = |
Ca2+ + |
CO32- |
|
(-x) |
x |
x |
Ksp = 4.5 x 10-9 = [Ca2+][CO32-] = [x][x]; x = (4.5 x 10-9)1/2 = 6.708 x 10-5 M
Sol = |_6.7 x 10-5 M______
Solubility = 6.708 x 10-5 mol/L x 100.09 g/mol = 6.714 x 10-3 g/L
100 mL = 0.100 L; 6.714 x 10-3 g/L x 0.100 L = 6.71 x 10-4 g / 100mL
10. The solubility product for CaF 2 is 4.9 x 10-11. Estimate the solubility of CaF 2 in
a. 0.50 M NaF à 0.50 M F- common ion; Na+ is a spectator ion
|
CaF2(s) = |
Ca2+ + |
2 F- |
|
(-x) |
x |
2x + 0.50 |
Ksp = [Ca2+][F-]2 = [x][2x + 0.50]2 ~ [x][0.50]2
x ~ 4.9 x 10-11 / (0.50)2 = 1.96 x 10-10 M
Sol. = |_2.0 x 10-10 M________
b. 0.25 M CaCl2 à 0.25 M Ca2+; Cl- is a spectator ion.
|
CaF2(s) = |
Ca2+ + |
2 F- |
xKsp = [Ca2+][F-]2 = [x + 0.25][2x]2 ~ [.25][2x]2 ~ .25 x 4 x2
x ~ (4.9 x 10-11 / (1.00)1/2 = 7.0 x 10-6 M
Sol. = |_7.0 x 10-6 M_________
Expect both common ion situations to be less soluble than in pure water (2.3 x 10-4 M) but expect the effect of the F- ion to be greater because of the 2 to 1 situation and the squaring associated with this situation.