# Acids and Bases Review

Chem I | Chem II AP
1)
A 6.15 g sample of benzoic acid, HC7H5O2, is ionized in enough water to make 600.0 mL of solution. Calculate the pH of the solution.
pKa (benzoic acid) = 4.2
6.15 g HC7H5O2 x
 1 mol 122 g
= 0.050 mol
 0.0500 mol 0.6000 L
= 0.840 M
 HC7H5O2 ↔ H+ + C7H5O2- Ka = 10-pKa = 10-4.2 = 6.31 x 10-5 0.0840 - x x x
Ka =
 [H+][C7H5O2-] [HC7H5O2]
6.31 x 10-5 =
 x2 0.0840
x2 = 5.30 x 10-6  x = 2.30 x 10-3
 pH = -log [H3O+] = -log (2.30 x 10-3) = 2.64
2)
The pH of a 0.10 M HNO3 solution is 2.4. Calculate the value of Ka at this temperature.
Give your answer in scientific notation either by using x10 between the number and exponent or by using e in place of the x10. For example 0.01 could be represented as 1x10-2 or 1e-2.
 [H3O+] = 10-pH = 10-2.4 = 3.98 x 10-3
 HNO2 + H2O ↔ H3O+ + NO2- 0.10 3.98 x 10-3 3.98 x 10-3
Ka =
 [H3O+][NO2-] [HNO2]
=
 (3.98 x 10-3)2 0.10
= 1.58 x 10-4
3)
Given: 0.10 M H2A solution. Ka1 = 1.0 x 10-5; Ka2 = 1.0 x 10-10
Find the [H3O+], [HA-], and [A-2]

[H3O+] =
[HA-] =
[A-2] =
 H2A + H2O ↔ H3O+ + HA- 0.10 - x1 x1 x1 assuming x1 is much smaller than 0.10:
Ka1 =
 [H3O+][HA-] [H2A]
1.0 x 10-5 =
 (x1)2 0.10
x1 = 1 x 10-3
 HA- + H2O ↔ H3O+ + A-2 1 x 10-3 - x2 1 x 10-3 + x2 x2 assuming x2 is much smaller than 0.001:
Ka2 =
 [H3O+][A-2] [HA-]
1.0 x 10-10 =
 (1 x 10-3)(x2) 1 x 10-3
x2 = 1 x 10-10
 x1 = [H3O+] = [HA-] = 1 x 10-3 x2 = [A-2] = 1 x 10-10
4)
Codeine, a cough suppressant extracted from crude opium, is a weak base with a Kb value of 1.6 x 10-6. What will be the pH of a 0.02 M solution of codeine?
 Cod + H2O ↔ OH- + HCod+ 0.02 - x x x
Kb =
 [HCod+][OH-] [Cod]
1.6 x 10-6 =
 x2 0.02
x2 = 3.2 x 10-8  x = 1.8 x 10-4
 pOH = -log [OH-] = -log (1.8 x 10-4) = 3.75
 pH = 14 - pOH = 14 - 3.75 = 10.25

Not knowing the specific chemical formula of Codeine will not stop you from working this problem. I indicated it in the above equations as Cod, but you can use any shorthand you wish. Because it says that Codeine is a weak base, I know that when it reacts with water it will form OH- in solution which leaves Codeine to accept the leftover proton from water.

5) Find the pH of each of the following 0.10 M solutions:
HNO2

pH =
 HNO2 + H2O ↔ H3O+ + NO2- 0.10 x x
Ka =
 [H3O+][NO2-] [HNO2]
4.5 x 10-4 =
 x2 0.10
x2 = 4.5 x 10-5  x = 6.76 x 10-3
 pH = -log [H3O+] = -log (6.76 x 10-3) = 2.17
KNO2

pH =
 NO2- + H2O ↔ OH- + HNO2 0.10 x x
Kb =
 [OH-][HNO2] [NO2-]
2.19 x 10-11 =
 x2 0.10
x2 = 2.19 x 10-12  x = 1.48 x 10-6
 pOH = -log [OH-] = -log (1.48 x 10-6) = 5.83 pH = 14 - 5.33 = 8.67

Because this salt comes from a weak acid and a strong base, it will make a base when hydrolyzed (reacted with water).

KOH

pH =
 KOH is a strong base and will completely ionize, therefore the concentration of the base is the same as the concentration of the [OH-].
 pOH = -log [OH-] = -log (0.10) = 1 pH = 13
NH4NO3

pH =
 NH4+ + H2O ↔ H3O+ + NH3 0.10 x x
Ka =
 [H3O+][NH3] [NH3]
5.6 x 10-10 =
 x2 0.10
x2 = 5.6 x 10-11  x = 7.45 x 10-6
 pH = -log [H3O+] = -log (7.45 x 10-6) = 5.13

Because this salt comes from a weak base and a strong acid, it will make an acid when hydrolyzed (reacted with water).

HCl

pH =
 HCl is a strong acid and will completely ionize, therefore the concentration of the acid is the same as the concentration of the [H3O+].
 pH = -log [H3O+] = -log (0.10) = 1
H2SO4

pH =
 H2SO4 is a strong acid and will completely ionize, therefore the concentration of the acid is the same as the concentration of the [H3O+].
 pH = -log [H3O+] = -log (0.10) = 1
Ca(OH)2

pH =

Ca(OH)2 is a strong base and will completely ionize but, since this base will break into a Ca+2 ion and 2 OH- ions, the the concentration of the base is half as much as the concentration of the [OH-].

 pOH = -log [OH-] = -log (0.20) = 0.70 pH = 13.3
HNO3

pH =
 HNO3 is a strong acid and will completely ionize, therefore the concentration of the acid is the same as the concentration of the [H3O+].
 pH = -log [H3O+] = -log (0.10) = 1
6) Complete the following and label conjugate acid/base pairs: In the blank under each molecule/ion indicate its identity using A, B, CA or CB
 NH3 + H2O → ↔ + CA
 HNO3 + H2O → ↔ + CB
 HF + H2O → ↔ + CB
 HNO2 + H2O → ↔ + CB
7) Find the pH given the following:
a)
pOH = 3.4

 pH = 14 - pOH = 14 - 3.4 = 10.6
b)
[OH-] = 6.3 x 10-3

 pOH = -log [OH-] = -log (6.3 x 10-3) = 2.20
 pH = 14 - pOH = 14 - 2.20 = 11.80
c)
[H3O+] = 8.7 x 10-4

 pH = -log [H3O+] = -log (8.7 x 10-4) = 3.06
8)
Which acid is stronger: Acid X with a pKa of 3.4 or Acid Z with a pKa of 5.2?
 Ka for Acid X = 10-pKa = 10-3.4 = 3.98 x 10-4
 Ka for Acid Z = 10-pKa = 10-5.2 = 6.31 x 10-6

A higher Ka value (smaller exponent) means you have a stronger acid, therefore X is a stronger acid than Z.