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CHEM 1000 A and V Fourth Test March 13, 2009
Part A. Answer all questions (5 marks each)
1. Which of vitamin A or vitamin B-6 would be more soluble in water? Why?
Vitamin B-6 is more soluble in water because of the large number of –OH groups in its structure, which
make H-bonds with water. Vitamin A is relatively non-polar, and so is less soluble.
Vitamin A
Vitamin B-6
2. Name or give the formula of a salt that could be used together with carbonic acid, H2CO3(aq), to make an
acidic buffer system.
Any salt containing the conjugate base of H2CO3, which is the bicarbonate ion, HCO3-. Some examples are
NaHCO3, KHCO3, Ca(HCO3)2, etc.
3. For H2CO3(aq), Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11. Why is Ka2 so much smaller than Ka1?
For this (and any polyprotic acid), the second hydrolysis constant is always smaller than the first because in
the second hydrolysis, a proton is being pulled away from a doubly charged anion (CO3-2 in this case),
whereas in the first hydrolysis, it is being pulled away from only a singly charged anion (HCO3- in this case).
4. Will the solubility of sodium carbonate, Na2CO3, increase or decrease as the pH is lowered? Why?
Kb for NaOH is very large. Ka for HCO3‾ is 5.6 x 10-11.
Na2CO3 + H2O(l) Ý 2 Na+(aq) + CO3-2(aq)
Na+(aq) + H2O(l)  no reaction, since the Na+ is the (weak) conjugate acid of a very strong base
CO3-2(aq) + H2O(l) Ý HCO3-(aq) + OH-(aq), since CO3-2 is the conjugate base of a weak acid.
Lowering the pH decreases [OH-(aq)]. This shifts the third reaction to the right, decreasing [CO3-2(aq)]. This
causes more Na2CO3 to dissolve to replace the CO3-2(aq). In other words, the solubility increases.
5.
Why is the vapour pressure of a solution of a non-volatile solute lower than that of the pure solvent?
Solute molecules at the surface “block” solvent molecules from entering the gas phase, lowering the partial
pressure of the solvent.
OR
The entropy of s solution is higher than that of the pure solvent. The positive entropy change upon
vapourization is therefore smaller in a solution than in a pure solvent. A smaller entropy change is less
favourable, hence less vapourization occurs.
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6. Why is HBr a stronger acid than HCl?
The H-Cl bond is stronger because the Cl is more electronegative than the Br. A stronger bond means a
weaker acid.
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 molality of a 4.07 M aqueous solution of NaBr. The solution has a density of 1.311 g cm-3.
4.07 M 
4.07 mol NaBr
1 L solution
1 L solution  1000 cm 3  1.311 g cm 3  1311 g solution
4.07 mol NaBr  (23.0  79.9) g mol 1  418.8 g NaBr
thus, mass H 2O  1311 418.8  892 g H 2O
m NaBr 
4.07 mol NaBr
 4.56 m
0.892 kg H 2 O
(b) Calculate the freezing point (oC) of a 2.2 m solution of FeCl3(aq). (For water, Kf = 1.86oC kg mol-1.)
Note that FeCl3  Fe+3(aq) + 3 Cl-(aq), thus m = 4(2.2) = 8.8 m
Tf = Kf m = 1.86oC kg mol-1(8.8 m) = 16.4oC
Thus, Tf = 0.0 – 16.4 = -16.4oC
(c) Calculate the vapour pressure (mm Hg) of a solution made by dissolving 4.3 moles of CoCl2(s) in 2.5 L of
water at 25oC. (The vapour pressure of pure water at 25oC is 23.8 mm Hg. Assume water has a density of
1.00 g cm-3.)
4.3 moles CoCl2  3(4.3) = 12.9 moles ions
n H 2 O  2.5 L 
X solute 
1000 cm3 1.00 g 1 mol


 138.9 mol
1L
1 cm3 18.0 g
n solute
12.9

 0.0849
n solute  n solvent 12.9  138.9
X solvent  1 X solute  1 0.0849  0.915
psolution  p *solvent (X solvent )  23.8 mm Hg(0.915)  21.78 mm Hg
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2. An aqueous solution contains 0.2 M methylamine, CH3NH2 (Kb = 3.7 x 10-4) and 0.25 M methylamine
hydrochloride, CH3NH3Cl.
(a) Calculate the pH of the solution.
K a (CH3 NH3Cl) 
Kw
K b(CH3 NH2 )

1.0  10 14
 2.70  10 11
4
3.7  10
pK a   log10 (K a )   log10 (2.70  10 11)  10.57
[acid]  [CH 3 NH 3 Cl]  0.25 M
[base]  [CH 3 NH 2 ]  0.20 M
 [acid] 
 0.25 
pH  pK a  log10 
  10.47
  10.57  log10 
 0.20 
 [base] 
(b) 0.010 mol of HCl are added to 1 L of the buffered solution from part (a). Write the reaction that occurs,
and calculate the resultant pH.
CH3NH2(aq) + HCl(aq)  CH3NH3+(aq) + Cl-(aq)
In 1.0 L of the buffer, we had 0.20 mol L-1 x 1 L = 0.20 mol of CH3NH2. Adding 0.010 mol HCl will
destroy 0.010 mol of it, lowering it to 0.019 mol. 0.019 mol / 1 L = 0.019 M
In 1.0 L of the buffer, we had 0.25 mol L-1 x 1 L = 0.25 mol of CH3NH3+. Adding 0.010 mol of HCl will
create another 0.010 mol of it, increasing it to 0.25 + 0.010 = 0.26 mol. 0.026 mol / 1 L = 0.026 M
 [acid] 
 0.26 
H  pK a  log10 
  10.43
  10.57  log10 
 0.19 
 [base] 
3.
Vanadium (V) metal crystallizes in a body centred cubic structure. It has an atomic radius of 134 pm.
Calculate the density (in g/cm3) of vanadium metal.
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d
l
l
m
l
Referring to the cubic unit cell above, m2 = l2 + l2
and, d2 = l2 + m2 = 3(l2)
In a body centred cubic structure, we know that the body diagonal, d, is equal to 4r, where r is the radius
of the atom.
4r 4(134  1012 m)
Thus, l 

 3.09  1010 m  3.09  108 cm
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The volume of the unit cell is therefore V  l3  (3.09 108 cm)3  2.96 1023 cm3
There are the equivalent of two atoms in a BCC unit cell. The mass of two vanadium atoms is
2  50.94 g mol1
mass 
 1.69  10 22 g
23
1
6.02  10 mol
mass
1.69  10 22 g
density 

 5.72 g cm3
23
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volume 2.96  10 cm
This problem can also be answered by noting that the space occupied in a BCC structure is 68%:
Volume of an atom = 4/3 πr3. Using r = 134 pm, V = 1.01 x 10-23 cm3
Volume occupied by two atoms in the unit cell is 2 (1.01 x 10-23 cm3) = 2.02 x 10-23 cm3
Volume of the unit cell is therefore 2.02 x 10-23 cm3 / 0.68 = 2.96 x 10-23 cm3
Mass of the unit cell is 1.69 x 10-23 g (as above)
Thus, the density is:
mass
1.69  1022 g
density 

 5.72 g cm3
23
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volume 2.96  10 cm
4.
Calculate the pH and the concentrations of all species (including OH-(aq)) in a 2 M solution of sulphurous
acid, H2SO3(aq). For this acid, Ka1 = 0.015 and Ka2 = 6.3 x 10-8.
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H2SO3(aq) + H2O(l) Ý H3O+(aq) + HSO3-(aq)
Ka1 = 0.015
H2SO3(aq) H3O(aq) HSO3-(aq)
2.0
0
0
Initial, M
-x
+x
+x
Change, M
2.0 - x
x
x
Equilibrium, M
Thus at equilibrium,
[H 3 O  ][HSO3  ]
x(x)

 0.015
[H 2SO3 ]
2. 0  x
x 2  0.015(2.0  x)
x 2  0.030  0.015 x
x 2  0.015 x  0.030  0
0.015  0.015 2  4(1)(0.030)
2(1)
0.015  0.347

 0.166 or  0.181
2
x
Using the positive root, x = 0.166
Thus, [H2SO3(aq)] = 2.0 – 0.166 = 1.834 M
[H3O+(aq)] = x = 0.166 M
[HSO3-(aq)] = x = 0.166 M
pH = -log10[H3O+(aq)] = -log10(0.166) = 0.780
[OH  (aq) ] 
Kw
1.0  10 14

 6.02  10 14 M

[H 3 O (aq) ]
0.166
HSO3-(aq) + H2O(l) Ý H3O+(aq) + SO3-2(aq)
Ka1 = 6.3 x 10-8
[H3 O ][SO3 2 ]
 K a2
[HSO3  ]
But from the first step, [H3O+(aq)] = [HSO3-(aq)]. Thus, [SO3-2(aq)] = Ka2 = 6.3 x 10-8 M