Mr. Storie
40S Chemistry
Acids Bases – Unit Review
ANSWER KEY
1. Explain how the Lewis definition of a base is different from that of the Arrhenius definition of a base. Are
Arrhenius bases also bases under the Lewis definition? Explain.
Lewis defined acids and bases as electron acceptors/donators respectively.
Arrhenius defined acids and bases based on the presence of H ion / OH ion in solution.
Lewis’ definition is the most broad, therefore all previous species of bases, whether defined by
Arrhenius or Bronsted-Lowry would also be Lewis bases.
2. Determine the type of reaction occurring below. If possible label the conjugate acid-base pair in the
appropriate place (Think about what is taking place in the reaction).
a. H2SO4 + H2O  HSO3-1 + H3O+
A
B
CB
CA
- Acid Dissociation reaction
b. Ca(OH)2 + 2 HNO3  Ca(NO3)2 + 2 H2O
B
A
Salt
- Neutralization reaction
c. NaCl  Na+ + ClSalt
- Decomposition reaction
3. What’s the pH of a 3.3 x 10-5 M NaOH solution?
Strong base – completely dissociates – [NaOH] = [OH-]
pOH = -log[OH-] = -log[3.3 x 10-5] = 4.48
pH = 14 – pOH
pH = 9.5
4. Find the pH of a 0.0050 M acetic acid solution. Ka = 1.8 x 10-5
CH3COOH 
[I]
0.0050
[C]
-x
[E]
0.0050-x
CH3COO- + H+
0
0
+x
+x
x
x
Ka = [CH3COO-][ H+]
[CH3COOH]
1.8x 10 -5 = [x]2
[0.0050]
1.8x 10 -5 (0.0050) = x2
√9.0 x 10 -8 = √x2
pH = -log[H+] = -log[0.0003] = 3.5
(Ignore –x, it’s too small)
x = 0.0003
Mr. Storie
Acids Bases – Unit Review
40S Chemistry
5. What is the pH of a solution that contains 2.4 x 10-5 moles of hydrobromic acid in 0.50 L of water?
2.4 x 10-5 moles
0.50 L
= 4.8 x 10 -5M
Strong acid – completely dissociates – [HBr] = [H+]
pH = -log[H+] = -log[4.8 x 10 -5M] = 4.3
6. What is the pH of a solution that contains 25 moles of nitric acid dissolved in 5000 L of water?
25 moles
5000 L
= 0.005 M
Strong acid – completely dissociates – [HNO3] = [H+]
pH = -log[H+] = -log[0.005 M] = 2.3
7. In a few steps, describe how you would titrate a base of unknown concentration with an acid with
concentration of 1 M.
Fill burette with unknown base – record initial volume.
Add a measured volume of Acid into a beaker.
Add an appropriate indicator into the beaker.
Slowly titrate by adding base from burette into beaker until a colour change (endpoint) is seen.
Record final volume of base.
Preform necessary calculations.
8.
I did a titration where it took 50 mL of 0.1 M hydrochloric acid to neutralize 500 mL of a base with
unknown concentration. Using this information, what was the concentration of the base?
*assume a 1:1 reaction:
1 HCl + 1 BOH  BCl + H2O
Convert mL into L: 50mlL = 0.05 L
500 mL = 0.5 L
0.1 moles HCl x 0.05 L x 1 mol BOH x
1L
1 mol HCl
1__ = 0.01 M BOH
0.5 L
Mr. Storie
Acids Bases – Unit Review
40S Chemistry
9. I did a titration where it took 25 mL of 5 M NaOH to neutralize 1000 mL of an acid with unknown
concentration. Using this information, what was the concentration of acid?
*assume a 1:1 reaction:
1 HX + 1 NaOH  NaX + H2O
Convert mL into L: 25 mL = 0.025 L
1000 mL = 1.0 L
5.0 moles NaOH x 0.025 L x 1 mol HX x
1L
1 mol NaOH
1__ = 0.125 M HX
1.0 L
10. If it takes 560 mL of 0.0050 M NaOH to neutralize 100.0 mL of H2SO4 solution with unknown
concentration, what was the original pH of the H2SO4 solution? (1.9)
1 H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
Convert mL into L: 560 mL = 0.56 L
100.0 mL = 0.1000 L
0.0050 moles NaOH
1L
x 0.56 L
x 1 mol H2SO4
2 mol NaOH
x
1__ = 0.014 M H2SO4
0.1000 L
Strong acid – completely dissociates – [H2SO4] = [H+]
*Only one H+ dissociates in original pH: H2SO4  HSO4- + H+
pH = -log[H+] = -log[0.014 M] = 1.9