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CHEM 1000 A and V Fourth Test March 14, 2008
Part A. Answer all questions (5 marks each)
1. Why do we have to consume larger amounts of vitamins B and C than of vitamins A, D, E and K?
Vitamins B and C are water soluble because they contain polar O-H groups. They are thus passed out of the
body more easily than the relatively non-polar, hence non-soluble vitamins A, D, E and K.
2. Why is HClO2 a stronger acid than HClO? Be specific.
The additional O atom in HClO2 results in greater attraction of the electrons towards the oxygens, hence
away from the O-H bond. This weakens the O-H bond and makes the acid stronger.
3. Describe how an acid-base indicator works.
If the indicator molecule is represented by HIn, then in water the reaction is HIn + H2O = H3O+ + In-. Thus
in low pH solutions, this equilibrium is shifted to the left, resulting in more HIn, and less In-. Since these
two molecules have different colors, the color of the solution changes with pH.
4. Will the solubility of ammonium sulphate, (NH4)2SO4, increase or decrease as the pH is lowered? Why?
pKb for NH3 is 1.8 x 10-5. pKa for HSO4‾ is 0.012.
NH3 is a weak base so NH4+ is a relatively strong acid. HSO4- is a fairly strong acid, so SO4-2 is a relatively
weak base. Thus, a solution of (NH4)2SO4(aq) is acidic, and the only relevant reaction is NH4+ + H2O = NH3
+ H3O+. Decreasing the pH will shift this reaction to the left, raising [NH4+]. This will cause less
(NH4)2SO4(s) to dissolve to make up for the increase in [NH4+]. In other words the solubility will decrease.
5. Why is hexane (C6H12) not very soluble in water?
Compounds A and B are usually only soluble in each other if the intermolecular forces A-A, B-B and A-B
are similar. In this case, water has relatively strong H-bonds, hexane has relatively weak dispersion forces,
so the forces are not the same. Another way of saying this is “like dissolves like”.
6. Name a salt that could be used to make a buffered system with the weak acid HF.
Any soluble salt that contains the F- ion, e.g. NaF(s), KF(s), etc. but not NH4F, since the ammonium ion is
acidic in solution.
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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 molality of a 3.74 M aqueous solution of KCl. The solution has a density of 1.16 g cm-3.
3.74 M 
3.74 mol KCl
1L solution
1L solution  1000 cm3  1.16 g cm 3  1,160 g solution
3.74 mol KCl  (39.1  35.5) g mol 1  279.0 g KCl
thus, mass H2O  1160  279  881g H2O
mKCl 
3.74 mol KCl
 4.25 m
0.881kg H2O
(b) Calculate the freezing point (oC) of a 3.2 m solution of MgCl2(aq). (For water, Kf = 1.86oC kg mol-1.)
Note that MgCl2  Mg+2(aq) + 2 Cl-(aq), thus m = 3(3.2) = 9.6 m
Tf = Kf m = 1.86oC kg mol-1(9.6 m) = 17.9oC
Thus, Tf = 0.0 – 17.9 = -17.9oC
(c) Calculate the vapor pressure (mm Hg) of a solution made by dissolving 2.2 moles of LiCl(s) in 4.4 L of water
at 25oC. (The vapor pressure of pure water at 25oC is 23.8 mm Hg. Assume water has a density of 1.00 g cm3
.)
2.2 moles LiCl  4.4 moles ions
nH2O
1000 cm3 1.00 g 1mol
 4.4 L 


 244.4 mol
1L
1cm3 18.0 g
Xsolute 
nsolute
4.4

 0.0179
nsolute  nsolvent 4.4  244.4
Xsolvent  1  Xsolute  1  0.0179  0.982
psolution  p * solvent (Xsolvent )  23.8 mm Hg(0.982)  23.38 mm Hg
2. A solution contains 0.2 M lactic acid, HC3H5O3(aq), and 0.15 M sodium lactate, NaC3H5O3(aq).
Ka for lactic acid = 1.4 x 10-4.
(a) Calculate the pH of the solution.
pKa = -log10(Ka) = -log10(1.4 x 10-4) = 3.85
[acid] = [H3C3H5O3] = 0.2 M
[base] = [H2C3H5O3-] = 0.15 M
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 [acid] 
pH  pK a  log10 

 [base] 
 0.20 M 
 3.85  log10 

 0.15 M 
 3.73
(b) 0.010 mol of NaOH is added to 1 L of the buffered solution from part (a). Write the reaction that occurs,
and calculate the resultant pH.
Some of the acid is destroyed (and an equivalent amount of base is created) according to:
H3C3H5O3(aq) + OH-(aq)  H2C3H5O3-(aq) + H2O(l)
Originally, the number of moles of acid was 0.2 mol/L x 1.00 L = 0.2 mol
Thus, after the addition of base, the number of moles of acid decreases by 0.01, and the acid
0.20 mol  0.01mol
concentration becomes
 0.19 M
1L
Similarly, according to the above reaction, the number of moles of base increases by 0.01, and its
0.15 mol  0.01mol
concentration becomes
 0.16 M
1L
Thus,
 [acid] 
pH  pK a  log10 

 [base] 
 0.19 M 
 3.85  log10 

 0.16 M 
 3.78
3.
Calculate the pH and concentrations of all species present (except water) in a 1.00 M aqueous
solution of the triprotic arsenic acid, H3AsO4aq). 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) Ý H3O+(aq) + H2AsO4-(aq)
K a1 
[H2 AsO4 (aq) ][H3O(aq) ]
[H3 AsO4(aq) ]
Thus, at equilibrium,
 5.6  103
x(x)
 5.6  10 3
1.0  x
Solving (must use the quadratic equation here since Ka > 10-3), we find x = 0.0722 M
Thus, [H3AsO4(aq)] = 1.0 – x = 1.0 – 0.0722 = 0.928 M
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[H2AsO4-(aq)] = x = 0.0722 M
[H3O+(aq)] = x = 0.0722 M
Kw
1014

[OH (aq) ] 

 1.39  10 13 M

[H3O (aq) ] 0.0722
pH = -log10[H3O+(aq)] = -log10(0.0722) = 1.14
H2AsO4-(aq) + H2O(l) Ý H3O+(aq) + HAsO4-2(aq)
K a2 
K a2 
[HAsO 4 2(aq) ][H3O  (aq) ]
[H2 AsO 4

(aq)
]
[HAsO 4 2(aq) ](0.0722)
(0.0722)
[HAsO 4
2
(aq)
 1.7  10 7
 1.7  10 7
]  1.7  10 7
HAsO4-2(aq) + H2O(l) Ý H3O+(aq) + AsO4-3(aq)
K a3 
[AsO4 3(aq) ][H3O  (aq) ]
[HAsO4
3
[AsO4 ] 
2
(aq)
]
 4.0  10 12
4.0  10 12 [HAsO 4 2(aq) ]
[H3O (aq) ]
4.0  10 12 (1.7  10 7 )
0.0722
 9.42  10 18 M

4.
Sodium (Na) metal crystallizes in a body centred cubic lattice and has a density of 0.971 g cm-3.
Calculate the atomic radius of sodium (in pm).
d
l
l
m
l
Referring to the cubic unit cell above, m2 = l2 + l2
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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. Thus, r 
3l2
.
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There are the equivalent of two atoms in a BCC unit cell. The mass of two sodium atoms is
mass 
2  22.99 g mol1
 7.64  10 23 g
6.02  1023 mol1
volume 
mass
7.64  10 23 g

 7.87  10 23 cm3
density 0.971g cm3
Since this is a cube, the edge length, l, is
l  3 volume
 3 7.87  10 23 cm3  4.28  10 8 cm
3(4.28  108 cm)2
3l2
r

 1.86  108 cm  186 pm
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This problem can also be answered by noting that the space occupied in a BCC structure is 68%:
1
1

 1.030 cm3g1
3
 0.971g cm
Volume per mole of Na metal = 1.030 cm3 g-1 x 23.0 g/mol = 23.69 cm3/mol
Volume per mole of Na atoms = 0.68 x 23.69 cm3/mol = 16.11 cm3/mol Na atoms
Volume per Na atom = 16.11 cm3 / mol / 6.02 x 1023 mol-1 = 2.68 x 10-23 cm3
Volume of a sphere = 4/3 πr3. Solving, r = 1.86 x 10-8 cm = 186 pm