1
CHEM 1000 A and V Fourth Test March 5, 2010
Part A. Answer all questions (5 marks each)
1.
Which of HNO2 and HClO2 is the stronger acid and why?
HClO2 is a stronger acid. The Cl atom is more electronegative than the N atom. The Cl atom therefore draws
electrons towards itself, reducing the electron density in the O–H bond, weakening this bond and making the acid
stronger.
2. Explain what happens when a small amount of NaOH(s) is added to an acidic buffer system composed of HNO2(aq) and
NaNO2(aq), and why the pH does not change much.
The NaOH destroys some of the acid:
NaOH(aq) + HNO2(aq) → H2O(l) + Na+(aq) + NO2–(aq)
but the equilibrium reaction HNO2(aq) + H2O(l) Ý H3O+(aq) + NO2– shifts, replacing most of the acid and restoring the
pH to close to its initial value.
3. What is the conjugate base of OH–(aq) ?
O2–(aq)
4. Will the solubility of sodium phosphate, Na3PO4, increase or decrease as the pH is increased? Why?
Kb for NaOH is very large. Ka for HPO42– is 4.8 x 10-13.
In water, sodium phosphate dissociates: Na3PO4(s) → 3 Na+(aq) + PO4–(aq).
Because HPO4–(aq) has a small Ka value, then PO43–(aq) is a relatively strong base. Thus the phosphate ion hydrolyzes
according to: PO4–(aq) + H2O(l) Ý HPO4–(aq) + OH–(aq)
Increasing the pH increases [OH–(aq)]. The equilibrium reaction thus shifts to the left to reduce the [OH–(aq)]. This
increases [PO43–(aq)]. As a result, more Na3PO4(s) crystallizes out of solution. The solubility therefore decreases.
5.
Describe in terms of the entropy and free energy changes of vapourization why the vapour pressure of a solution is
lower than that of the pure solvent.
Adding a solute increases the entropy of the solution. The difference in entropy between the solution and the gas
phase is therefore smaller with a solution than with a pure solvent. Thus Svap,solution < Svap,solvent. For the
vapourization process, Gvap = Hvap – TSvap. If Svap is smaller, then Gvap is more positive, which means less
evaporation, i.e. a lower vapour pressure above a solution.
6. Why is adding HCl to water an exothermic process?
The very strong attraction between the H+ or Cl– ions and water molecules result in the formation of ion–dipole
complexes (H+(aq) and Cl–(aq)). Formation of these “bonds” releases energy.
Part B. Answer three of the following four questions (B1 to B4, 20 marks each). If you answer all four, the best
three will be used to calculate your mark for part B.
2
B1. (a) [8] Calculate the molarity of CaCl2 in a 2.25 m aqueous solution of CaCl2. The solution density is 1.311 g cm-3.
2.25 m CaCl2 means 2.25 moles CaCl2 in 1 kg water.
2.25 moles CaCl2 x (40.1 + 2(35.45)) g/mol = 249.8 g CaCl2
The mass of the solution is 1,000 g + 249.8 g = 1,250 g
The solution volume is therefore 1,250 g / 1.311 g mL–1 = 953.3 mL
Molarity = moles/L = 2.25 moles /0.9533 L = 2.36 M
(b) [6] The freezing point (oC) of an aqueous salt solution is –11.5 oC. Calculate the boiling point of this same solution.
(For water, Kf = 1.86 oC kg mol-1 and Kb = 0.51 oC kg mol-1)
T f  K f m
Tb  K b m
 T f
 Kb 
 K
 f


11.5 oC
o
1 

0.51
C
kg
mol
 3.15o C


o
1 
 1.86 C kg mol 

Thus, Tb = 100.00 + 3.15 = 103.15oC
(c) [6] Calculate the vapour pressure (Pa) of a solution made by dissolving 2.3 moles of Na2SO4(s) in 0.5 L of water at
25oC. (The vapour pressure of pure water at 25oC is 3173 Pa. Assume water has a density of 1.00 g cm-3.)
p  p o solv X solv
2.3 moles Na2SO4 dissociates into 4.6 moles Na+(aq) and 2.3 moles SO42–(aq), i.e. 6.9 moles total ions
0.5 L water = 500 mL x 1 g/mL = 500 g water
500 g water / 18.02 g/mol = 27.75 mol water
27.75 mol
 0.801
27.75 mol  6.9 mol
p  0.801(3173 Pa )  2541 Pa
X solv 
B2. An aqueous solution contains 0.5 M hydrazine, H2NNH2 (Kb = 1.3 x 10-6) and 0.25 M hydrazine hydrochloride,
H2NNH3Cl.
(a) [10] Calculate the pH of the solution.
The H2NNH3Cl will dissociate in water to give H2NNH3+ and Cl–.
The acid in this solution is H2NNH3+. We need pKa for this species. Note that it is the conjugate acid of H2NNH2, and
for conjugates, pKa = 14–pKb (or Ka = Kw/Kb).
Thus, pKa = 14–(–log10(Kb)) = 14 + log10(1.3 x 10–6) = 8.11
3
 [acid ] 
pH  pK a  log10 

 [base] 
 [ H NNH 3 ] 
 pK a  log10  2

 [ H 2 NNH 2 ] 
 [0.25 M ] 
 8.11  log10 

 [0.50 M ] 
 8.41
(b) [10] 0.010 mol of HNO3 (Ka = 30) are added to 1 L of the buffered solution from part (a). Write the reaction that
occurs, and calculate the resultant pH.
Adding the acid destroys some of the base according to:
H2NNH2 + HNO3 → H2NNH3+ + NO3– (The net ionic equation is also fine: H2NNH2 + H3O+ → H2NNH3+ + H2O,
or even H2NNH2 + H+ → H2NNH3+)
We can therefore calculate the new concentrations after this reaction (but before any re–equilibration takes place)
using an ice table. Note that the HNO3 is the limiting reagent:
Initial
Change
Equilibrium
H2NNH2
0.5 M x 1 L = 0.5 mol
–0.01 mol
0.49 mol
HNO3
0.01 mol
–0.01 mol
0.00 mol
H2NNH3+
0.25 M x 1 L = 0.25 mol
+0.01 mol
0.26 mol
The we use the Henderson–Hasslebalch equation to determine the new pH:
 [ H NNH 2 ] 
pH  pK a  log10  2
 
 [ H 2 NNH 3 ] 
 [0.26 M ] 
 8.11  log10 

 [0.49 M ] 
 8.39
B3. (a) [10] Ksp for BaF2 is 1.8 x 10–7. Calculate the solubility of BaF2(s) (in g/L) in water.
BaF2(s) Ý Ba2+(aq) + 2 F–(aq)
Ksp = 1.8 x 10–7 = [Ba2+(aq)][F–(aq)]2
In a solution containing only barium fluoride, [Ba2+(aq)] = x and [F–(aq)] = 2x
Thus, x(2x)2 = 1.8 x 10–7
4x3 = 1.8 x 10–7
x3 = 4.5 x 10–8
x = 3.56 x 10–3
Since x = [Ba2+] and since 1 mol BaF2 dissolves to give 1 mole Ba2+, then the solubility of BaF2 = x moles/L.
4
3.56 x 10–3 mol/L x (137.3 + 2(19.00)) g/mol = 0.624 g/L
(b) [5] Calculate the solubility of BaF2 (in g/L) in 0.01 M NaF(aq).
The NaF is very soluble, so 0.01 M NaF has [F–(aq)] = 0.01 M
1.8 107
 1.8 103 mol / L
[ F  ]2
0.012
 (137.3  2(19.00)) g / mol  0.32 g / L
[ Ba 2 ] 
K sp

(c) [5] Does the solubility of BaF2(s) increase or decrease as the pH is increased? Show reactions to support your
answer.
BaF2(s) Ý Ba2+(aq) + 2 F–(aq)
But the fluoride ion is the conjugate base of a weak acid (HF), so the solution is basic according to:
F–(aq) + H2O(l) Ý HF(aq) + OH–(aq)
Increasing the pH increases [OH–(aq)], so the last equilibrium shifts left, which increases [F–(aq)]. This causes the
first equilibrium to shift left, effectively decreasing the solubility of BaF 2(s).
B4. Calculate the pH and the concentrations of all species (except water) in a 2 M solution of chromic acid,
H2CrO4(aq). For this acid, Ka1 = 0.18 and Ka2 = 3.2 x 10-7.
H2CrO4(aq) + H2O(l) Ý HCrO4–(aq) + H3O+(aq)
[ HCrO4  ( aq ) ][ H 3O  ( aq ) ]
[ H 2CrO4( aq ) ]
 K a1  0.18
x2
 0.18
2 x
x 2  0.18 x  0.36  0
here, a = 1, b = 0.18 and c =  0.36
x
b  b 2  4ac
 0.517 or  0.697
2a
We must use the positive root because x is a concentration. Thus, x = 0.517
[H3O+(aq)] = x = 0.517 M [4]
[HCrO4–(aq)] = x = 0.517 M [4]
[H2CrO4(aq)] = 2 – x = 1.483 M [4]
pH = –log10[H3O+(aq)] = –log10(0.517) = 0.29 [4]
HCrO4–(aq) + H2O(l) Ý + H3O+(aq) + CrO42–(aq)
5
[CrO4
2
( aq )
][ H 3O
[ HCrO4

[CrO4 2 ( aq ) ]x
x
[CrO4
2
( aq )

( aq ) ]
( aq )
]
 K a2  3.2 107
 3.2 107
]  3.2 107 M  4