18.1 Lewis Acids and Bases KEY
1. AlCl3 reacts with chloride ions to form the complex ion AlCl4-.
a. Draw Lewis structures to illustrate the interaction between AlCl3 and the chloride ion to form
the complex ion AlCl4-.
b. Explain how the behavior of AlCl3 is consistent with the definition of a Lewis acid.
Electron pair acceptor – accepts non-bonded pair from Clc. What is the name for the special type of covalent bond that exists between AlCl3 and the
chloride ion?
Covalent/Coordinate/Dative bond
2. For each of the following species, state whether it is most likely to act as a Lewis acid or a Lewis
base:
a. PH3
Lewis Base
b. BCl3
Lewis Acid
c. H2S
Lewis Base
d. Cu2+
Lewis Acid
3. Which equation represents an acid-base reaction according to the Lewis theory but not the
Brønsted-Lowry theory?
A.
NH3 + HCl
B.
C.
NH4Cl
2H2O
H3O+ + OH–
NaOH + HCl
NaCl + H2O
D.
CrCl3 + 6NH3
[Cr(NH3)6]3+ + 3Cl–
4. Which statement explains why ammonia can act as a Lewis base?
A. Ammonia can donate a lone pair of electrons.
B. Ammonia can accept a lone pair of electrons.
C. Ammonia can donate a proton.
D. Ammonia can accept a proton.
5. Define an acid in terms of the Lewis theory. Deduce, giving a reason, whether NF3 is able to
function as a Lewis acid or as a Lewis base.
A Lewis acid can accept a pair of electrons;
it is a Lewis base as it can donate the lone/non-bonding pair of electrons (on the N atom);
18.2 Calculations Involving Acids and Bases KEY
6. For each of the following, determine [H+], [OH-] , pH and pOH
a. 1.5 M HF (Ka = 7.0 X 10 –4)
R HF
↔
H+
+
FI 1.5
0.0
0.0
C -x
+x
+x
E 1.5
x
x
𝑲𝒂 =
[𝑯+ ][𝑭− ]
[𝑯𝑭]
𝒙𝟐
𝟕. 𝟎𝒙𝟏𝟎−𝟒 = 𝟏.𝟓
√𝟏. 𝟎𝟓𝒙𝟏𝟎−𝟑 = √𝒙𝟐
0.032
pH = -log(0.032) = 1.49
pOH = 12.51
[OH-] = 3.09x10-13
b. 1.0 M CH3COOH ( Ka =1.8 X 10 –5)
R CH3COOH ↔
H+
+
I 1.0
0.0
C -x
+x
E 1.0
x
𝑲𝒂 =
[𝑯+ ][𝑪𝑯𝟑𝑪𝑶𝑶− ]
[𝑪𝑯𝟑𝑪𝑶𝑶𝑯]
CH3COO0.0
+x
x
𝒙𝟐
𝟏. 𝟖𝒙𝟏𝟎−𝟓 = 𝟏.𝟎
√𝟏. 𝟎𝟓𝒙𝟏𝟎−𝟑 = √𝒙𝟐
0.032
pH = -log(0.032) = 1.49
pOH = 12.51
[OH-] = 3.09x10-13
7. The pH of a 0.1 mol dm-3 solution of cyanic acid, HCN, is 3.
a. Write the equation for the ionization of cyanic acid.
HCN
↔
H+
+
CNb. Write the Ka expression.
[𝑯+ ][𝑪𝑵− ]
𝑲𝒂 =
[𝑯𝑪𝑵]
c. Calculate Ka.
R HCN ↔
I 0.1
C -0.001
E 0.1
𝑲𝒂 =
H+
+
0.0
+0.001
0.001
CN0.0
+0.001
0.001
[𝑯+ ][𝑪𝑵− ] 𝟎. 𝟎𝟎𝟏𝟐
=
= 𝟏. 𝟎𝒙𝟏𝟎−𝟓
[𝑯𝑪𝑵]
𝟎. 𝟏
pH = 3 = 10-3 = 0.001 (H+ conc)
8. Determine the pH of a 1.2 mol dm-3 solution of HF (Ka = 7.2 x 10-4). (REPEAT TO 6a)
R HF
↔
H+
+
FI 1.2
0.0
0.0
C -x
+x
+x
E 1.2
x
x
𝑲𝒂 =
[𝑯+ ][𝑭− ]
𝒙𝟐
𝟕. 𝟐𝒙𝟏𝟎−𝟒 = 𝟏.𝟐
[𝑯𝑭]
√𝟖. 𝟔𝒙𝟏𝟎−𝟒 = √𝒙𝟐
0.029
pH = -log(0.029) = 1.53
9. Determine the pOH and the pH of a 1.8 mol dm-3 solution of KF that has a pKa 3.17.
Ka = 10-pKb = 10-3.17 = 6.76 x 10-4
R
I
C
E
HF
1.8
-x
1.8
𝑲𝒂 =
+
↔
H2 O
-
[𝑯𝟑𝑶+ ][𝑭− ]
[𝑯𝑭]
H3 O+ +
0.0
+x
+x
𝒙𝟐
𝟔. 𝟕𝟔𝒙𝟏𝟎−𝟒 = 𝟏.𝟖
F0.0
+x
+x
√𝟏. 𝟐𝒙𝟏𝟎−𝟑 = √𝒙𝟐
0.0348
pH = -log(0.0348) = 1.46
pOH = 12.54
10. The pKb for a base is 5. What is the pH of a 0.1 mol dm-3 solution of the base?
Kb = 10-pKb = 10-5 = 1.0x10-5
R
I
C
E
B
0.1
-x
0.1
𝑲𝒃 =
+
H2 O
-
[𝑩𝑯+ ][𝑶𝑯− ]
[𝑩]
↔
BH+
0.0
+x
+x
𝒙𝟐
𝟏. 𝟎𝒙𝟏𝟎−𝟓 = 𝟎.𝟏
+
OH0.0
+x
+x
√𝟏. 𝟎𝒙𝟏𝟎−𝟔 = √𝒙𝟐
0.001
pOH = -log(0.001) = 3
pH = 11
11. A 0.75 mol dm-3 solution of a weak acid, HX, has a pH of 1.35. Determine Ka for the acid.
pH = 1.35 = 10-1.35 = 0.045
R
I
C
E
HX ↔
0.75
-0.045
0.705
𝑲𝒂 =
[𝑯+ ][𝑿− ]
[𝑯𝑿]
H+
+
0.0
+0.045
0.045
=
𝟎.𝟎𝟎𝟒𝟓𝟐
𝟎.𝟕𝟎𝟓
X0.0
+0.045
0.045
= 𝟎. 𝟎𝟎𝟐𝟖
12. The Ka value for an acid is 1.0 × 10–2. What is the Kb value for its conjugate base?
a. 1.0 × 10-2
b. 1.0 × 10-6
c. 1.0 × 10-10
d. 1.0 × 10-12
13. Four aqueous solutions, I, II, III and IV, are listed below. What is the correct order of increasing pH
of these solutions?
i. 0.100 mol dm-3 HCl
-log(0.100) = 1
-3
ii. 0.010 mol dm HCl
-log(0.010) = 2
iii. 0.100 mol dm-3 NaOH
-log(0.100) = 1
14-1 = 13
iv. 0.010 mol dm-3 NaOH
-log (0.010) = 2
14-2 = 12
a. I, II, III, IV
b. I, II, IV, III
c. II, I, III, IV
d. II, I, IV, III
14. Propanoic acid, CH3CH2COOH is a weak acid.
(a) Give the equation for the ionization of propanoic acid in water and deduce the
expression for the ionization constant, Ka, of propanoic acid.
C2H5COOH
𝑲𝒂 =
+
H2 O
↔
C2H5COO-
+
H+
[𝑪𝟐𝑯𝟓𝑪𝑶𝑶− ][𝑯+ ]
[𝑪𝟐𝑯𝟓𝑪𝑶𝑶𝑯][𝑯𝟐𝑶]
(b) Calculate the Ka value of propanoic acid using the pKa value in the Data Booklet.
Ka = 10-pKa = 10-4.87 = 1.35x10-5
(c)
Use your answer from (b) to calculate the [H+] in an aqueous solution of propanoic acid
of concentration 0.0500 mol dm–3, and hence the pH of this solution.
𝒙𝟐
𝟏. 𝟑𝟓𝒙𝟏𝟎−𝟓 = 𝟎.𝟎𝟓𝟎𝟎 = 𝟖. 𝟐𝟐𝒙𝟏𝟎−𝟒
15. (a)
(i)
Write the equation for the reaction of ammonia with water.
NH3
(ii)
pH = -log(8.22x10-4) = 3.09
+
H2 O
↔
NH4+ +
Derive the expression for Kb for this reaction.
𝑲𝒃 =
[𝑵𝑯𝟒+ ][𝑶𝑯− ]
[𝑵𝑯𝟑][𝑯𝟐𝑶]
OH-
(b) Using information from Table 21 in the Data Booklet, determine the pOH of a
0.20 mol dm–3 solution of ammonia.
pKb = 4.75 = 10-4.75 = 1.78x10-5
𝒙𝟐
𝟏. 𝟕𝟖𝒙𝟏𝟎−𝟓 = 𝟎.𝟐𝟎 = 𝟎. 𝟎𝟎𝟐
pOH = -log(0.002) = 2.72
16. Determine the pH of the solution resulting when 100 cm3 of 0.50 mol dm–3 HCl(aq) is mixed with 200
cm3 of 0.10 mol dm–3 NaOH(aq).
𝑴=
𝒎𝒐𝒍𝒔
𝒅𝒎𝟑
𝒙
𝑯𝑪𝒍 𝟎. 𝟓𝟎 = 𝟎.𝟏𝟎𝟎 = 𝟎. 𝟎𝟓 𝒎𝒐𝒍𝒔
𝑵𝒂𝑶𝑯 𝟎. 𝟏𝟎 =
𝒙
= 𝟎. 𝟎𝟐 𝒎𝒐𝒍𝒔
𝟎. 𝟐𝟎𝟎
HCl remaining (0.05 – 0.02) = 0.03 mols HCl
Total volume = 300 cm3 solution
𝟎.𝟎𝟑
New [HCl] = 𝟎.𝟑𝟎𝟎 = 𝟎. 𝟏𝟎 𝒎𝒐𝒍𝒔 𝒅𝒎−𝟑
-log(0.10) = 1
18.3 pH Curves
17. When the following 1.0 mol dm-3 solutions are arranged in order of increasing pH, which is the
correct order?
i. Ammonium chloride
<7 weak base with strong Alkali (cation hydrolysis)
ii. Ammonium ethanoate
~7 weak acid/weak base
iii. Sodium ethanoate
>7 weak acid (anion hydrolysis)
a. I, II, III
b. II, I, III
c. III, I, II
d. III, II, I
18. Which mixture would produce a buffer solution when dissolved in 1.0 dm3 of water?
a. 0.50 mol of CH3COOH and 0.50 mol of NaOH
b. 0.50 mol of CH3COOH and 0.25 mol of NaOH
c. 0.50 mol of CH3COOH and 1.00 mol of NaOH
d. 0.50 mol of CH3COOH and 0.25 mol of Ba(OH)2 (does not work because 2 OH- released per mol)
19. A buffer solution can be made by dissolving 0.25 g of sodium ethanoate in 200 cm3 of 0.10 mol dm–3
ethanoic acid. Assume that the change in volume is negligible.
a)
Define the term buffer solution.
A solution that resists small changes in acidity/alkalinity
b)
Calculate the concentration of the sodium ethanoate in mol dm–3.
𝟎. 𝟐𝟓𝒈 𝑵𝒂𝑪𝑯𝟑𝑪𝑶𝑶 𝟎. 𝟎𝟎𝟑 𝒎𝒐𝒍𝒆𝒔
=
= 𝟎. 𝟎𝟏𝟓 𝒎𝒐𝒍𝒔 𝒅𝒎−𝟑
𝟖𝟐. 𝟎𝟑𝟒𝒈
𝟎. 𝟐 𝒅𝒎𝟑
c)
Calculate the pH of the resulting buffer solution by using information from Table 21 of
the Data Booklet.
[𝒔𝒂𝒍𝒕]
𝒑𝑯 = 𝒑𝑲𝒂 + 𝒍𝒐𝒈 [𝒂𝒄𝒊𝒅] = 𝟒. 𝟕𝟔 + 𝐥𝐨𝐠
𝟎.𝟎𝟏𝟓
𝟎.𝟏𝟎
= 𝟑. 𝟗𝟒
20. Which substances could be added to a solution of ethanoic acid to prepare an acidic buffer
solution?
I. Hydrochloric acid
II. Sodium ethanoate
salt of weak base with strong alkali
III. Sodium hydroxide strong base
A.
B.
I and II only
I and III only
C.
II and III only
D.
I, II and III
21. Calculate the pH of a buffer solution made by reacting 20 cm3 0.10 mol dm-3 HCl(aq) with 40
cm3 0.10 mol dm-3 NH3(aq) at 298K. The pKb for ammonia is 4.75.
Same concentration
𝒑𝑶𝑯 = 𝒑𝑲𝒃 + 𝒍𝒐𝒈
[𝒔𝒂𝒍𝒕]
𝟎. 𝟏𝟎
= 𝟒. 𝟕𝟓 + 𝐥𝐨𝐠
= 𝟒. 𝟕𝟓
[𝒃𝒂𝒔𝒆]
𝟎. 𝟏𝟎
14-4.75 = pH = 9.25
22. (a)
(i)
Calculate the Ka value of methanoic acid, HCOOH, using table 21
Ka = 10-pKa = 10-3.75 = 1.78x10-4
(ii)
Based on its Ka value, state and explain whether methanoic acid is a strong or
weak acid.
Weak  small Ka value means less dissociation
(iii) Calculate the hydrogen ion concentration and the pH of a 0.010 mol dm–3
methanoic acid solution. State one assumption made in arriving at your answer.
R
I
C
E
HCOOH
0.010
-x
0.010
𝑲𝒂 =
[𝑯𝑪𝑶𝑶− ][𝑯+ ]
[𝑯𝑪𝑶𝑶𝑯]
↔
H+
0.0
+x
x
HCOO0.0
+x
x
+
𝒙𝟐
𝟏. 𝟕𝟖𝒙𝟏𝟎−𝟒 = 𝟎.𝟎𝟏𝟎 1.33x10-3
pH = -log(1.33x10-3) = 2.87
assumed [HCOOH]i = [HCOOH]eqm
(b) Explain how you would prepare a buffer solution of pH 3.75 starting with methanoic
acid.
1. add strong base (NaOH) or alkali (NaHCOO) to methanoic acid
2. react with enoung to have equimolar amounts (1/2 acid – ½ salt)
3. pH = pKa  3.75 for HCOOH
23.
(i)
25.0 cm3 of 1.00 × 10–2 mol dm–3 hydrochloric acid solution is added to 50.0 cm3 of
1.00 × 10–2 mol dm–3 aqueous ammonia solution. Calculate the concentrations of
both ammonia and ammonium ions in the resulting solution and hence determine the pH
of the solution.
Moles:
𝒙
𝑵𝑯𝟑 = 𝟎. 𝟎𝟏 = 𝟎.𝟎𝟓 = 𝟓. 𝟎𝟎𝒙𝟏𝟎−𝟒
𝒙
𝑯𝑪𝒍 = 𝟎. 𝟎𝟏 = 𝟎.𝟎𝟐𝟓 = 𝟐. 𝟓𝒙𝟏𝟎−𝟒
R NH3
+
HCl

NH4+ +
Cl- (dissociated form)
I 5.00x10-4
2.5x10-4
0.0
-
C -2.5x10-4
-2.5x10-4
+2.5x10-4
-
E 2.5x10-4
0.0
2.5x10-4
-
[𝒔𝒂𝒍𝒕]
𝟐.𝟓𝒙𝟏𝟎−𝟒
𝒑𝑶𝑯 = 𝒑𝑲𝒃 + 𝒍𝒐𝒈 [𝒃𝒂𝒔𝒆] = 𝟒. 𝟕𝟓 + 𝐥𝐨𝐠 𝟐.𝟓𝒙𝟏𝟎−𝟒 = 𝟒. 𝟕𝟓
14-4.75 = pH = 9.25
(ii)
State what is meant by a buffer solution and explain how the solution in (i), which
contains ammonium chloride dissolved in aqueous ammonia, can function as a buffer
solution.
Buffer resists change in pH due to small amounts of acid/base addition
NH3 + H+ ↔ NH4+
NH4+ + OH- ↔ NH3 + H2O
24. For each of the following salts, tell if the aqueous solution is acidic, basic, or neutral. When acidic or
basic, show the hydrolysis equation.
KF, KCl, NH4NO3, NaClO4, Mg(NO3)2
KF
Strong alkali with weak acid  basic
K+ + H2O ↔ KOH + H+ (no rxn KOH dissociates)
F- + H2O ↔ HF + OH- BASIC
KCl
Strong alkali and strong acid  neutral
K+ + H2O ↔ KOH + H+ (no rxn KOH dissociates)
Cl- + H2O ↔ HCl + OH- (no rxm HCl dissociates)
NH4NO3 weak base with strong acid  acidic
NH4+ + H2O ↔ NH3 + H3O+ ACIDIC
NO3- + H2O ↔ HNO3 + OH- (no rxn HNO3 dissociates)
NaClO4
strong alkali with weak acid  basic
Na+ + H2O ↔ NaOH + H+ (no rxn NaOH dissociates)
ClO4- + H2O ↔ HClO4 + OH- BASIC
Mg(NO3)2
weak alkali with strong acid  acidic
Mg2+ + H2O ↔ Mg(OH)2 + H+
ACIDIC
NO3 + H2O ↔ HNO3 + OH (no rxn HNO3 dissociates)
25. Which compound will dissolve in water to give a solution with a pH greater than 7?
a. Sodium chloride (SA + SB) pH = 7
b. Potassium carbonate
(SB + WA)
c. Ammonium nitrate (WB + SA) pH < 7
d. Lithium sulfate (SA + SB) pH = 7
26. Deduce whether the pH of the resulting salt solution will be greater than, less than or equal to 7 when
the following solutions exactly neutralize each other.
a. H2SO4(aq) + NH3(aq)
< 7 (strong acid/weak base – cation hydrolysis)
b. H3PO4(aq) + KOH(aq)
> 7 (strong base/weak acid – anion hydrolysis)
c. HNO3(aq) + Ba(OH)2(aq)
= 7 (strong acid and base)
27. 50 mL of a base with unknown concentration was titrated with 1.5M HCl. Determine the
concentration of the base if it required 25 mL of acid to neutralize the base. (Think how a dilution calc
is set up)
MAVA = MBVB
(1.5)(25) = MB(50)
= 0.75 mols dm-3 of base
28. Sketch titration curves for:
a. The titration of a strong acid and a weak base (base
added to acid)
b. The titration of a strong acid and a strong base
(base added to acid)
pH = 7
pH < 7
c. The titration of a weak acid and a strong base (base added
to acid)
d. The titration of a weak acid and a weak base (base added
to acid)
pH > 7
pH ~ 7
29. Hydrochloric acid (in the flask) is to be titrated with aqueous sodium carbonate (in the burette).
a. Would you choose methyl orange or phenolphthalein for this titration? Explain. (use data
booklet)
Methyl Orange – strong acid+ weak base  buffer region made, but pH at equivalence
is below 7
b. What color change would you expect to see at the end point?
Red  yellow
Acid  ‘base’
c. Explain why adding too much indicator could lead to an inaccurate titration result.
Indicators are weak acids therefore base needs to neturalize indicator as well. Too
much indicator leads to inaccurate amount of base needed for true neutralization.
d. The laboratory has run out of both methyl orange and phenolphthalein. Below are listed
some indicators that are available. Which would you use to replace your original choice?
Explain.
Indicator
bromophenol blue
enough to methyl orange.
pKa
4.0
Color Change
yellow to blue – pH = pKa, close
bromothymol blue
7.0
yellow to blue
thymol blue
8.9
yellow to blue
30. Separate 20.0 cm3 solutions of a weak acid and a strong acid of the same concentration are
titrated with NaOH solution. Which will be the same for these two titrations?
I.
Initial pH – no (weak increases, strong decreases)
II. pH at equivalence point – no (weak > 7, strong = 7)
III. Volume of NaOH required to reach the equivalence point – yes, as long as acid is
same concentrations
A.
I only
B.
III only
C.
D.
I and II only
II and III only
31. Which curve is produced by the titration of a 0.1 mol dm−3 weak base with 0.1 mol dm−3 strong
acid?
12
A.
12
B.
8
8
pH
pH
4
4
Volume of titrant
12
C.
Volume of titrant
12
D.
8
8
pH
pH
4
4
Volume of titrant
32.
(a)
Volume of titrant
Explain why a 1.0 mol dm–3 solution of sodium hydroxide has a pH of 14 whereas
1.0 mol dm–3 ammonia solution has a pH of about 12. Use equations in your answer.
NaOH  strong base – completely dissociates – high [OH-], low [H+] – high pH =
14
NaOH  Na+ + OHNH3  weak base – partially dissociates – lower [OH-], higher [H+] – lower pH =
12
NH3 ↔ NH4+ + OH-
(b)
20.0 cm3 of a known concentration of sodium hydroxide is titrated with a solution of
nitric acid. The graph for this titration is given below.
13
pH
7
0
(i)
10
20
Volume of nitric acid / cm 3
30
State an equation for the reaction between sodium hydroxide and nitric acid.
NaOH (aq) + HNO3 (aq)  NaNO3 (aq) + H2O (l)
(ii)
Calculate the concentration of the sodium hydroxide solution before the titration.
pH = 13
14 – 13 = 1 (pOH)
dissociation, [NaOH] = [OH-]
10-1 = 1.0x10-1 [OH-], because of full
(iii) From the graph determine the volume of nitric acid required to neutralize the
sodium hydroxide and calculate the concentration of the nitric acid.
18.0 cm3 required for neturalization
MAVA = MBVB
MA(18.0) = (0.1)(20.0)
= 0.11 mols dm-3 of acid
(iv) Predict the volume of ethanoic acid of the same concentration as the nitric acid in
(b) (iii), required to neutralize 20.0 cm3 of this sodium hydroxide solution.
18.0 cm3 – due to the buffering action of a weak acid and strong base – the
point of inflection is at the same point
33. Which statement about indicators is always correct?
a. The mid-point of an indicator’s color change is at pH=7
b. The pH range is greater for indicators with higher pKa values
c. The color red indicates an acidic solution
d. The pKa values of an indicator is within its pH range
34. Bromocresol green has a pH range of 3.8-5.4 and changes color from yellow to blue as the pH
increases
a. Of the four types of titrations shown, state in which two of these this indicator could be
used.
Strong Acid – Strong Base
Strong acid – weak base
b. Suggest a value for the pKa of this indicator
4.6 – midpoint of indicator range
c. What color will the indicator be at pH 3.6?
yellow