1
Name:_______________________________________
Student Number:____________________
CHEM 1000 Fourth Test March 4,2005
Make sure this test has six pages! Authorized memoranda: Calculator only
Part A. Answer all questions (5 marks each)
1. Why does spreading salt on the roads at -30oC not melt the ice?
Even saturated sodium chloride solutions freeze at a higher temperature than -30oC.
2. Name or give the formula of a salt that could be used to make an acidic buffer solution with aqueous iodic acid, HIO3.
We could use any salt of this weak acid, for instance NaIO3, KIO3, etc.
3. Will a solution of NaNO2(aq) be acidic, basic or neutral? Why? (Ka for HNO2(aq) is 4.5 x 10-4)
Since HNO2(aq) is a weak acid, its conjugate the nitrite ion, NO2-(aq), will be a base. The solution is therefore basic.
4. List all species present in an aqueous solution of the triprotic arsenic acid H3AsO4(aq).
H3AsO4(aq), H2AsO4¯(aq), HAsO4¯(aq), AsO4¯(aq), H3O+(aq) and OH¯(aq) (as well as water of course).
5. Why is HCl(aq) a stronger acid than HF(aq)?
F is more electronegative than Cl. Thus, the H-F bond is more ionic than the H-Cl bond. The more ionic bond is stronger,
resulting in a weaker acid.
6. CaF2(s) has a Ksp value of 1.8 x 10-7. As the pH is lowered, will the solubility of this salt increase, decrease, or remain
unchanged? Why? (Ca(OH)2 is strong base, HF is a weak acid)
The CaF2(s) dissolves according to CaF2(s)  Ca+2(aq) + 2 F¯(aq). However, since HF is a weak acid, F¯(aq) must be a base, and
so the following equilibrium also exists: F¯(aq) + H2O(l)  HF(aq) + OH¯(aq). As the pH is lowered, [OH¯(aq)] decreases.
The second equilibrium therefore shifts right to replace it, consuming some of the F ¯(aq). The first equilibrium therefore
shifts right to replace the F¯(aq), i.e. the solubility increases.
Part B. Answer three of the following four questions (20 marks each). If you answer all four, the best three will
be used to calculate your mark for part B.
1. A solution is made by dissolving 0.50 moles of FeCl2(s) in 1.00 kg of water. The density of the solution is 1.0 g/mL.
[4] (a) Calculate the freezing point (oC) of the solution. (For water, Kf = 1.86oC kg mol-1)
0.50 mol
 3  1.50 m (since FeCl2  Fe+2(aq) + 2 Cl-(aq))
1.00 kg
Tf  K f m
m
 1.86C kg mol 1  1.50 mol kg 1
 2.79C
Thus, Tf  0.00C  2.79C  2.79C
[4] (b) Calculate the boiling point (oC) of the solution. (For water, Kb = 0.51oC kg mol-1.)
2
Tb  K b m
 0.51C kg mol1 1.50 mol kg 1
 0.76C
Thus, Tb  100.00C  0.76C  100.76C
[4] (c) Calculate the osmotic pressure (atm) of the solution at 25oC. (R = 0.082 L atm K-1mol-1)
The solution volume is 1.00 L since the density is 1.0 g/mL.
  MRT
 1.50 mol 
1
1

  0.082 L atm K mol  (25  273) K
 1.00 L 
 36.7 atm
[4] (d) Calculate the mole fraction of water in this solution.
n solute  1.50
n solv 
1000 g
 55.5
18.0 g mol1
thus, X solv 
n solv
55.5

 0.974
n solv  n solute 55.5  1.50
[4] (e) Calculate the vapor pressure (mm Hg) of the solution at 25oC. (The vapor pressure of pure water at 25oC is
23.8 mm Hg.)
p  23.8 mm Hg  0.974  23.2 mm Hg
2. A buffered aqueous solution contains the base propylamine (C3H9N, Kb = 5.1 x 10-4) and the salt propylamine
hydrochloride, C3H9NHCl.
[8] (a) Calculate the pH of a solution which is 1.00 M propylamine and 2.00 M propylamine hydrochloride.
C3H9N(aq) + H2O(l)  C3H9NH+(aq) + OH¯(aq) Kb = 5.1 x 10-4
C3H9NHCl + H2O(l)  C3H9NH+(aq) + Cl¯(aq)
In this system, the acid is the C3H9NH+(aq). Ka for this acid is calculated from Kb of the propylamine:
Ka = Kw / Kb = 1 x 10-14 / 5.1 x 10-4 = 1.96 x 10-11
Thus, pKa = -log10(Ka) = -log10(1.96 x 10-11) = 10.7
[acid] = [C3H9NH+(aq)] = 2.00 M
[base] = [C3H9N(aq)] = 1.00 M
3
 [acid] 
pH  pK a  log10 

 [base] 
 2.00 M 
 10.7  log10 

 1.00 M 
 10.4
[6] (b) Calculate the concentration of propylamine hydrochloride required to give a pH of 10.0 in a solution of 2.00 M
propylamine.
Rearranging the Henderson Hasselbalch equation,
[acid]
 pK a  pH
[base]
[acid]
or,
 10(pK a  pH)
[base]
log10
or, [acid]  [base]  10(pK a  pH)
 2.00 M  10(10.7 10.0)
 10.02 M
[6] (c) 0.10 mol of HCl are added to 1 L of the buffered solution from part (a). Calculate the resultant pH.
First, the added acid titrates an equivalent amount of base according to C3H9N(aq) + H3O+(l)  C3H9NH+(aq) + H2O(l)
Thus, the acid concentration increases from to (2.00 + 0.10) = 2.10 M and the base concentration decreases to (1.00 0.10) = 0.90 M. The pH is found from the Henderson-Hasselbalch equation:
 [acid] 
pH  pK a  log10 

 [base] 
 2.10 M 
 10.7  log10 

 0.90 M 
 10.3
3. Calculate the pH and concentrations of all species present in a 1 M aqueous solution of the triprotic arsenic acid,
H3AsO4(aq). For this acid, Ka1 = 5.6 x 10-3, Ka2 = 1.7 x 10-7 and Ka3 = 4.0 x 10-12.
H3AsO4 (aq) + H2O(l) = H3O+(aq) + H2AsO4- (aq)
[H3O (aq) ][H 2 AsO4-(aq) ]
[H3AsO4(aq) ]
Initial, M
Change, M
Equilibrium, M
H3AsO4 (aq)
1.00
-x
1.00-x
 K a1  5.6 103
H3O+(aq)
0
+x
x
H2AsO4- (aq)
0
+x
x
4
2
Thus at equilibrium,
x
 5.6 103
1.00  x
Solving using the quadratic equation, x = 0.072
Thus, [H3O+(aq)] = [H2AsO4-(aq)] = x = 0.072 M
pH = -log10[H3O+(aq)] = -log10(0.072) = 1.14
[H3AsO4(aq)] = 1.00 – x = 0.928 M
[10]
The concentration of the second hydrolysis product, HAsO4-2, is found from:
H2AsO4-(aq) + H2O(l) = H3O+(aq) + HAsO4-2 (aq)
[H3O  (aq) ][HAsO4 -2(aq) ]
-
[H 2 AsO4 (aq) ]
x[HAsO4 -2(aq) ]
 K a 2  1.7  107
 1.7  107
x
[HAsO4 -2(aq) ]  1.7  107
[5]
And the concentration of the third hydrolysis product, AsO4-3 (aq) is found from:
HAsO4-2 (aq) + H2O(l) = H3O+(aq) + AsO4-3 (aq)
[H 3O  (aq) ][AsO 4 3(aq) ]
2
[HAsO 4 ]
or, [AsO 4 3(aq) ] 

 K a3  4.0  1012
K a3[HAsO 4 2 ]
[H 3O  (aq) ]
[5]
4.0  1012 (1.7  107 )
 9.44  1018 M
0.072
[20] 4. An element crystallizes in a body centred cubic lattice and the solid has a density of 10.22 g cm-3. The edge
length of its unit cell is 3.147 x 10-8 cm. Calculate the atomic mass of the element (18 marks) and predict which
element it might be (2 marks).
The volume of the unit cell = V = l3 = (3.147 x 10-8 cm)3 = 3.12 x 10-23 cm3
The mass of the unit cell = density x volume = 10.22 g cm-3 x 3.11 x 10-23 cm3 = 3.18 x 10-22 g
The unit cell in a body centred lattice contains 2 atoms. Thus, each atom has a mass of 3.18 x 10 -22 g / 2 = 1.59 x 10-22 g
A mole of atoms thus have a mass of 1.59 x 10-22 g x 6.02 x 1023 mol-1 = 95.9 g mol-1
Must be Mo (Molybdenum)