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Name:_______________________________________
Student Number:____________________
Chemistry 65.100 A and V Fourth Test March 7, 2003
Make sure this test has six pages! Authorized memoranda: Calculator only
Part A. Answer all questions (5 marks each)
1. Why is sodium chloride not effective at melting ice if the temperature is below -20.7°C?
Saturated sodium chloride freezes at this temperature.
2. Which of the two substances shown below is a stronger acid and why?
F3C
H2
C
C
H2
H2
C
O
C
H2
COH
6,6,6-trifluorohexanoic acid
F2HC
H2
C
C
H2
H2
C
O
C
H2
COH
6,6-difluorohexanoic acid
The 6,6,6-trifluorohexanoic acid is the stronger acid. The extra fluorine, being an electronegative atom, draws
electrons towards itself, withdrawing them from the O-H bond. This weakens the O-H bond causing the
compound to be more acidic..
3. Will an aqueous solution of NH4Br be acidic, basic or neutral? Why? (Ka for HBr(aq) is very large;
Kb for NH3 = 1.8 x 10-5)
Since HBr is a strong acid, its conjugate base (Br-) is a weak base, and will have no effect on the pH of the
solution. Since NH3 is a weak base, its conjugate acid is a relatively strong acid, and will cause the pH of the
solution to drop due to the reaction:
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq). The solution will therefore be acidic.
4. Will the solubility of sodium bicarbonate, NaHCO3(s), increase or decrease as the pH is lowered? Why?
The dissolution reaction of NaHCO3(s) is NaHCO3(s) + H2O(l)  Na+(aq) + HCO3-(aq)
The bicarbonate ion, being the conjugate base of the weak acid H2CO3(aq), hydrolyses according to
HCO3-(aq) + H2O(l)  H2CO3(aq) + OH-(aq)
Decreasing the pH decreases [OH-(aq)], shifting the hydrolysis reaction to the right, thus decreasing [HCO3-(aq)].
This will cause the dissolution reaction to shift to the right also, increasing the solubility of NaHCO3(aq).
5. What is the conjugate base of NH3?
NH26. Define the term “unit cell”.
A unit cell is the smallest unit in a crystalline solid that repeats in the x, y and z directions.
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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) Calculate the freezing point (oC) of a solution that contains 50 g NaCl per litre of water
(for water, Kf = 1.86oC kg mol-1. The density of water = 1.00 g/mL)
n NaCl 
50 g
 0.855 mol
58.5 g mol  1
nsolute  2  0.855  171
. mol (sin ce NaCl  Na  ( aq )  Cl  ( aq ) )
mass of water  1L 
msolute 
1000 mL 100
. g
1 kg


 100
. kg
1L
1 mL 1000 g
171
. mol
 171
. mol kg  1
100
. kg
 Tf  K f msolute
 186
.  C kg mol  1  171
. mol kg  1
 318
. C
Thus, Tf  0.00  318
.   318
. C
(b) Calculate the boiling point (oC) of the same solution. (For water, Kb = 0.51oC kg mol-1.)
 Tb  K b msolute
 0.51 C kg mol 1  171
. mol kg 1
 0.87 C
Thus, Tf  100.00  0.87  100.87 C
(c) Calculate the vapor pressure (mm Hg) of the solution at 25oC. (The vapor pressure of pure water at 25oC is
23.8 mm Hg.)
According to Raoult’s law, psoln = posolventXsolvent
nsolute  171
. mol
nsolvent 
1000 g
 555
. mol
18.0 g mol  1
Thus, Xsolvent 
nsolvent
555
.

 0.970
nsolvent  nsolute 555
.  171
.
Thus, pso ln  238
. mm Hg  0.970  231
. mmHg
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2. A buffered solution contains ammonia, NH3(aq) (Kb = 1.8 x 10-5) and ammonium chloride, NH4Cl(aq).
(a) Calculate the pH of a solution containing 1.00 M ammonia and 0.50 M ammonium chloride.
Here, [base] = [NH3(aq)] = 1.00 M
[acid] = [NH4+(aq)] = 0.50
Kw
10
.  10 14

.  10 10
Ka = Ka(NH4+(aq)] =
 5  556
K b ( NH 3( aq ) ) 18
.  10
Thus, pKa = -log10(Ka) = -log10(5.56 x 10-10) = 9.25
According to the Henderson-Hasselbalch equation:
[ acid ]
[ base]
0.50
 9.25  log10
100
.
 9.55
pH  pK a  log10
(b) 0.10 mol of HCl are added to 1 L of the buffered solution from part (a). Calculate the resultant pH.
The 0.10 mol of HCl will consume 0.10 mol of ammonia according to:
NH3(aq) + HCl(aq)  NH4+(aq) + Cl-(aq)
Thus, [NH3(aq)] drops to 1.00 – 0.10 = 0.90 M
And[NH4+(aq)] increases to 0.50 + 0.10 = 0.60
And so the pH can be calculated as:
[ acid ]
[ base]
0.60
 9.25  log10
0.90
 9.43
pH  pK a  log10
3. Calculate the pH and concentrations of all species present in a 1 M aqueous solution of the diprotic
oxalic acid, H2C2O4(aq). For this acid, Ka1 = 0.059, Ka2 = 6.4 x 10-5.
The acid first hydrolyses according to:
H2C2O4(aq) + H2O(l)  H3O+(aq) + HC2O4-(aq)
[ H 3O  ( aq ) ][ HC2 O4  ]
 K a1
For which the equilibrium expression is
[ H 2 C2 O4 ( aq ) ]
H2C2O4(aq) H3O+(aq)
Initial
Change
Equilibrium
1.00
-x
1.00 - x
0
+x
x
HC2O4-(aq)
0
+x
x
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Thus at equilibrium, we have
( x)( x)
 0.059
100
. x
Since the acid is relatively strong, we can not make the assumption that x << 1.00, and we must use
the quadratic equation.
x2 + 0.059x – 0.059 = 0
0.0592  4(1)(  0.059)
2(1)
 0.059  0.485

2
 0.213 or  0.272
Since x is a concentration, only the positive root makes any sense.
x
 0.059 
Thus, [H3O+(aq)] = [HC2O4-(aq)] = x = 0.213 M
And pH = -log10[H3O+(aq)] = -log10(0.213) = 0.67
The second hydrolysis is HC2O4-(aq)+ H2O(l)  H3O+(aq) + C2O4-2(aq), for which the equilibrium expression is
[ H 3O ( aq ) ][C2O4 2 ]
 Ka 2
[ HC2O4  ( aq ) ]
Rearranging to solve for the only unknown,
[ H 3O ( aq ) ]
[ C2 O 4 ] 
K
[ HC2O4  ( aq ) ] a 2
but from the first hydrolysis, [H3O+(aq)] = [HC2O4-(aq)] = 0.213 M
2
Thus, [C2O4-2(aq)] = Ka2 = 6.4 x 10-5 M
4. Iridium (Ir) metal crystallizes in a face centred cubic lattice and has a density of 22.5 g cm-3. Calculate
the atomic radius of Iridium (pm).
The mass of one Ir atom is
192.2 g mol  1
 319
.  10 22 g
6.02  1023 atomsmol  1
A face centred unit cell contains the equivalent of 4 atoms, and therefore has a mass of
4 x 3.19 x 10-22 g = 1.28 x 10-21g
The volume of the unit cell is therefore:
mass
128
.  10 21 g
Volume 

.  1023 cm3
3  567
density 22.5 g cm
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The length of one side of this cube is volume1/3 = (5.67 x 10-23cm3)1/3 = 3.84 x 10-8 cm
The diagonal of a face of the face centred cube is 4r. Thus, if the length of a side is l, then (4r)2 = l2 + l2,
or r = (l2/8)1/2 = (3.84 x 10-8 cm)2/8)1/2 = 1.36 x 10-8 cm = 136 pm