Acids and Bases Review

Chem I | Chem II AP
Your score: 0%
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
HC7H5O2H++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+H2OH3O++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]
Give all of your answers in scientific notation.
    
[H3O+] =
[HA-] =
[A-2] =
H2A + H2OH3O++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- + H2OH3O++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+H2OOH-+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+H2OH3O++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-+H2OOH-+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++H2OH3O++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.