Download Unit 6 Review Answers

Unit 6 Review Answers
Student Textbook pages 620–623
Answers to Knowledge/Understanding
Questions
16.
Multiple Choice
1. (c)
2. (b)
3. (a)
4. (d)
5. (d)
6. (d)
7. (c)
8. (d)
9. (c)
10. (e)
17.
18.
19.
Short Answer
11. (a) HCO3−(aq) + HPO42−(aq) ⇀
↽ CO32−(aq) + H2PO4−(aq)
(b) acid: HCO3−; conjugate base: CO32−
base: HPO42−; conjugate acid: H2PO4−
(c) Reactants are favoured: Ka(HCO3−) < Ka(H2PO4−)
12. pH = −log(0.004) = 2.4
13.
Concentration
(mol/L)
C6H5COOH(aq) +H2O(
)
C6H5CO−(aq) +H3O+(aq)
0.048
0
0
Change
0.048 – x
+x
+x
Equilibrium
0.048 – x
x
x
Initial
x )( x )
K a = (0.(048
= 6.3 × 10−5
− x)
Using the approximation method:
x = 1.7 × 10−3 mol/L
[H3O+] = 1.7 × 10−3 mol/L
pH = 2.76
[OH−] = 5.9 × 10−12
pOH = 11.23
14. Since acids and bases dissociate into ions in aqueous
solution, it would be reasonable to infer that acids
and bases are electrolytes.
15. (a) H2SO4 and NaOH are examples of an Arrhenius
acid and base.
(b) H2O and NH3 are examples of a Brønsted-Lowry
acid and base. Water can’t be an Arrhenius acid,
because it already exists as the aqueous solution in
which Arrhenius reactions occur. (Water could be
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MHR • Unit 6 Acids and Bases
considered a modernized Arrhenius acid, however.) Ammonia can’t be an Arrhenius base, because
it does not contain hydroxide ions. (Ammonia
could be considered a modernized Arrhenius base,
however.)
Students should disagree. A solution with a pH of 4
has [H3O+] = 0.0001 mol/L, whereas a solution with
a pH of 3 has [H3O+] = 0.001 mol/L.
The pH of pure (i.e., distilled) water is 7, because the
concentration of [H3O+] = [OH−] = 1.0 × 10−7 mol/L,
assuming a temperature of 25°C.
(a) Describing acids and bases as strong and weak
means that they are being compared in terms of
the extent to which they dissociate.
(b) Describing acids and bases as concentrated and
dilute means that they are being compared in
terms of the concentration of their dissociated
ions.
(a) HClO(aq) + H2O(l) ⇀
↽ ClO−(aq) + H3O+(aq)
Ka =
 ClO −  H O+ 

  3

[HClO]
(b) H2S(aq) + H2O(l) ⇀
↽ HS−(aq) + H3O+(aq)
Ka =
HS −  H O+ 

  3

[H2S]
20. HI(aq) > H3PO4(aq) > HF(aq) > CH3COOH(aq)
21. CH3COO−(aq) > F−(aq) > H2PO4−(aq) > I−(aq)
22. (a) The 0.01 mol/L solution has a higher pH,
because a pH of 2 is higher than a pH of 1.
(b) The 0.1 mol/L basic solution has a higher pH,
because bases have a pH > 7.
(c) The solution with a pOH of 7 has a higher pH,
because a pH of 8 is higher than a pH of 7.
(d) The solution with Ka = 4 × 10−5 has a higher pH,
because the lower Ka value means that the acid is
weaker, and a weaker acid has a higher pH.
23. (a) pH = −log(5.04 × 10−3) = 2.30
(b) The solution is acidic.
24. Amphoteric refers to the ability of a chemical species
to act as an acid or a base in a reaction. For example,
the hydrogen sulfide ion, HS−, is amphoteric, acting
as an acid in the first reaction below and as a base in
the second reaction. (In these reactions, water is also
amphoteric.)
HS-(aq) + H2O(l) ⇀
↽ H2S(aq) + OH−(aq)
HS-(aq) + H2O(l) ⇀
↽ S2− + H3O+(aq)
Answers to Inquiry Questions
25. (a)
(b)
Kb =
Kw
Ka
=
29. A titration curve for a polyprotic acid would show
1.0 × 10−14
= 1.6 × 10−5
6.2 × 10−10
Concentration
(mol/L)
CN−(aq)
OH−(aq)
+HCN(aq)
0.120
0
0
Change
0.120 – x
+x
+x
Equilibrium
0.120 – x
x
x
Initial
+H2O(
)
x )( x )
−5
K b = 0.(120
− x = 1.6 × 10
Using the approximation method:
x = 1.4 × 10−3 mol/L
pOH = 2.85
pH = 11.15
26. NaOH(aq) ⇀
↽ Na+(aq) + OH−(aq)
NH3(aq) + H2O(l) ⇀
↽ NH4+(aq) + OH−(aq)
CH3COO−(aq) + H2O(l) ⇀
↽ CH3COOH(aq) + OH−(aq)
CH3COOH(aq) + H2O(l) ⇀
↽ CH3COO−(aq) +
+
H3O (aq)
27. Working backwards from the titration:
From the titration,
2NaO(aq) + H2SO4(aq) ⇀
↽ 2H2O(l) + Na2SO4(aq)
number of moles of NaOH used in titration
= (0.230 mol/L)(0.0279 L) = 6.4 × 10−3 mol
number of moles of H2SO4
= 6.4 × 10−3 ÷ 2 = 3.2 × 10−3 mol
Number of moles of reacted H2SO4
= 0.0125 mol − 3.2 × 10−3 mol = 0.0093 mol
NH3(g) + H2SO4(aq) ⇀
↽ (NH4)2SO4(aq)
Number of moles of NH3 consumed in the above
reaction = 0.0093 mol × 2 = 0.0186 mol
For the initial reaction:
(NH4)2SO4(aq) + 2NaOH(aq) ⇀
↽
2NH3(g) + 2Na2SO4(aq) + H2O(l)
Number of moles of (NH4)2SO4 used to produce
0.019 mol NH3(g) = 0.0093 mol
Mass (NH4)2SO4 = 0.0093 mol × 132 g/mol = 1.23 g
% (NH4)2SO4 in initial sample
= (1.23 g ÷ 10.0 g) × 100% = 12.3%
28. See student textbook pages 608 and 609 for examples
of curves for a strong acid-strong base titration
(Figure 15.11) and a weak acid-strong base titration
(Figure 15.12). A strong base-weak acid titration
would have a large pH to start and a low pH to
finish. The equivalence point would appear above
pH = 7.
two equivalence points, one for each proton that is
dissociated.
30. (a) The concentration of a concentrated solution of a
strong acid will be different because 100% of the
acid molecule dissociates.
(b) The concentration of a dilute solution of a strong
acid will be different, because 100% of the acid
molecule will dissociate.
(c) The concentration of a concentrated solution of a
weak acid will be about the same, because very
little of the acid molecule dissociates.
(d) The concentration of a dilute solution of a weak
acid will be different, for the same reason as in (c).
31. [HAdiss] = 0.03 × 0.25 mol/L = 0.0075 mol/L
Concentration
(mol/L)
HA(aq)
+H2O(
)
A−(aq)
+H3O+(aq)
Initial
0.25
0
0
Change
–x
+x
+x
0.25 – x
x
x
Equilibrium
Assuming x is negligible:
[H3O+] = 0.0075 mol/L
pH = −log[H3O+] = −log(0.0075) = 2.12
pOH = 14.00 − 2.12 = 11.88
[OH−] = Kw/[H3O+]
= 1.0 × 10−14/7.5 × 10−3 = 1.3 × 10−12
Ka =
=
 A −  H O+ 

  3

[HA ]
0.0075
(
)(0.0075)
(0.25)
= 2.2 × 10−4
32. (a) i. yellow ii. blue iii. colourless
(b) pH is about 10.
33. Let x represent the concentration of H3O+ and
C6H5COO−.
[C6H5COOH]
= 0.25 −5 ≥ 500 ,
Ka
6.3 × 10
therefore you can neglect x in the denominator.
2
K a = [C H xCOOH]
6 5
x = (6.3 × 10−5 )(0.25)
x = 4.0 × 10−3
Therefore,
% dissociation =
4.0 × 10−3 mol/L
× 100%
0.25 mol/L
= 1.6%
Unit 6 Acids and Bases • MHR
231
34.
[ H 3O + ] = − log 3.44
= 3.6 × 10−4 mol / L
initial [HX] = 0.250 mol/0.655 L = 0.382 mol/L
[H3O + ][ X − ]
[HX]
(3.63 × 10−4 )2
=
(0.382 − 3.63 × 10−4 )
Ka =
= 3.45 × 10−7
35. Let x represent [H3O+]
H O+   Z − 
 3


Ka =
1.55 × 10−4 =



[HZ ]
x2
0
.
075
− x)
(
0 = x 2 + 1.55 × 10−4 x − 1.16 × 10−5
−1.55 × 10−4 ± (1.55 × 10−4 )2 − 4( −1.16 × 10−5 )
x=
2
x = 3.3 × 10−3
Therefore, [H3O+] = 3.3 × 10−3 mol/L
pH = −log(3.3 × 10−3) = 2.48
Ka =
1.55 × 10−4 =
H O +   Z − 
 3
 

[HZ ]
(
x2
0.045 − x
)
0 = x + 1.55 × 10−4 x − 16.98 × 10−6
2
−1.55 × 10−4 ± (1.55 × 10−4 )2 − 4( −6.98 × 10−6 )
2
x=
x = 2.6 × 10−3
pH = −log(2.6 × 10−3) = 2.6
pOH = 14.0 − 2.6 = 11.4
Answers to Communication Questions
36. The two acid solutions are not equally strong. The
hydronium ions in the hydrochloric acid solution dissociate completely (HCl is a strong acid), while the
hydronium ions in the acetic acid solution dissociate
only to a small extent.
37. Adding H3O+: NO2− + H3O+ ⇀
↽ HNO2 + H2O
+
− ⇀
Adding OH−: H3O + OH ↽ 2H2O
The hydroxide ion will react with the H3O+ formed
by the dissociation of HNO2.
38. (a) You would select an indicator that changes colour
over a pH that includes 7 at the equivalence
point.
(b) You would need to select an indicator that
changes colour over a pH range in which the
equivalence point is above 7.
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MHR • Unit 6 Acids and Bases
39. The end point signals the end of the titration, because
the indicator has changed colour. The equivalence
point occurs when the reactants have completely
reacted with each other. The end point may occur
either before or after the equivalence point, because it
is affected by the Ka of the titrated acid (and the Ka of
the indicator).
40. Students’ answers to this question should reflect personal experience, rather than “textbook description.”
Their descriptions should include references to pH,
end point, and equivalence point as they relate to the
colour change from clear to red (or pink).
41. n NaOH(aq) = 0.03893 L × 4.500 × 10–3 mol/L
= 0.1752 × 10–4 mol
The neutralization reaction for sodium hydroxide and
sulfuric acid is:
2NaOH(aq) + H2SO4(aq) → 2Na2SO4(aq) + H2O(l)
There are 2 mol NaOH for each mol H2SO4(aq).
0.1752 × 10–4 mol
= 0.8759 ×10 –5 mol
n H2SO4(aq) =
2
× 10 mol
= 3.504 × 10–3 mol
[H2SO4(aq)] = 8.759
0.025L
= 3.5 mol/L
+
42. (a) [NH4 ] = 4.12 mol/L; [Cl−] = 4.12 mol/L
(b) [Ba] = 0.275 mol/L; [OH2] = 0.550 mol/L
(c) [NH4+] = 0.543 mol/L × 3 = 1.629 mol/L;
[PO43−] = 0.543 mol/L
(d) [K+] = 0.704 mol/L; [ClO4−] = 0.704 mol/L
Note: In the first printing of the student textbook,
questions 43 and 44 below are equilibrium questions,
which may be assigned as further practice of skills
learned in Chapter 13, and applied in Unit 6.
43. Because the equilibrium constant is relatively large,
the percent of sulfur dioxide that reacts is also
relatively large. Twice as much SO2 reacts, compared
to O2.
–5
44.
mol
[NO2 ] = 0.4100
.00 L
= 0.0250 mol/L
At equilibrium, 26% of the NO2 remains.
[NO2] = 0.26 × 0.0250 mol/L = 6.5 × 10−3 mol/L
74% has reacted, and 2 mol NO2 form 1 mol N2O4.
mol/L
[N 2O4 ] = 0.74 × 0.0250
2
= 9.3 × 10−3 mol/L
K=
[ N2O 4 ]
[NO2 ]2
−3
= (9.3 × 102 )
[NO2 ]
=
(9.3 × 10−3 )
(6.5 × 10−3 mol/L )
= 220
= 2.2 × 102
45. (a) Students should expect it to be a base.
(b) Ca10(PO4)6(OH)2(s) + 8H+(aq) ⇀
↽
6CaHPO4(s) + 2H2O(l) + 4Ca2+(aq)
(c) Candy contains sucrose, which will remove
hydroxyapatite from teeth by reacting with it at
low pH to give Ca2+(aq) and H2PO4−(aq) that are in
soluble form.
NH 2 – (CH2)4 – NH2 NH 2 – (CH2)4 – NH3+
46.
OH–
0.10
0
0
change
–0.0021
+0.0021
+0.0021
equilibrium
0.0979
0.0021
0.0021
initial conc.
Kb =
(0.0021)2
= 4.5 × 10–5
–.0979
47. The conductivity of 0.1 mol/L HCl is higher than for
0.1 mol/L CH3COOH because the HCl is
completely ionized whereas the CH3COOH is only
partly ionized. For solutions of these two acids at a
concentration of 1 × 10–7 mol/L, the HCl is still
ionized completely and the CH3COOH again is only
partly ionized. However at such a low concentration
the conductivity of the HCl is the same as for water.
There will be no discernable difference in the
conductivities at this low concentration.
48. One drop of concentrated acid will have a high concentration of H3O+ ions. (E.g., For HCL [H3O+] =
about 12.5 mol/L, for H2SO4[H3O+] = about 35
mol/L. When one drop of acid is added to a bucket of
water, the H3O+ ion concentration is diluted by a
very large factor.
49. (a) A triprotic acid is an acid that has three protons
that may be dissociated, each with its own Ka
value.
(b) Stepwise Equation 1:
H3PO4(aq) + H2O(l) ⇀
↽ H2PO4−(aq) + H3O+(aq)
Stepwise Equation 2:
H2PO4−(aq) + H2O(l) ⇀
↽ HPO42−(aq) + H3O+(aq)
Stepwise Equation 3:
HPO42−(aq) + H2O(l) ⇀
↽ PO43−(aq) + H3O+(aq)
(c) H2PO4−(aq) acts as an acid in equation 2 above. In
the equation below, it acts as a base.
H2PO4−(aq) + H2O(l) ⇀
↽ H3PO4(aq) + OH−(aq)
(d) Students may say that H3PO4(aq) is the stronger
acid because it is higher in the list of relative
strengths of acids and bases. Some students may
suggest that H3PO4(aq) is a stronger acid because
its Ka value is larger than that of H2PO4−(aq),
which is a “stronger” answer.
50. Water could have a pH greater than 7.0 at a temperature different from 25°C.
51. (a) Students might suggest either of these equations:
NaHCO3(aq) → Na+(aq) + H+(aq) + CO3−(aq)
NaHCO3(aq) → Na+(aq) + H+(aq) + CO32−(aq)
(b) The litmus test indicates that the solution is basic,
so the solution contains a considerable [OH−].
Neither of these equations, written according to
the original Arrhenius theory, is consistent with
the litmus test. Equation two above even implies a
red litmus result! Accordingly, the litmus test contradicts the Arrhenius theory equations.
(c) A student could have two frames of thought.
Perspective I:
NaHCO3(aq) + H2O(l) →
Na+(aq) + H3O+(aq) + CO32−(aq)
or, without the spectator, Na+:
HCO3−(aq) + H2O(l) → H3O+(aq) + CO32−(aq)
Perspective II
NaHCO3(aq) + H2O(l) →
Na+(aq) + H2CO3(aq) + OH−(aq)
or, without the spectator, Na+:
HCO3−(aq) + H2O(l) → H2CO3(aq) + OH−(aq)
Note that, if a student writes equations according
to Perspective I, the student still has a set of
equations that contradict the litmus test. But if
the student then realizes that the hydrogen
carbonate ion is amphoteric, another valid set of
equations according to Perspective II may be
written. Given the litmus test, the student should
then analyze and evaluate the equations to state
that Perspective II is not only correct according to
modernized Arrhenius, but also consistent with
the litmus test. Some students will have realized
this out of part (b), and will present only
Perspective II.
Unit 6 Acids and Bases • MHR
233