Honors Chemistry
Ka, Kb and Neutralization Rxns.
1. Describe the difference between a weak acid and a strong acid. Give an example of each.

weak acids partially ionizes in solution, whereas strong acids completely ionize in solution. Weak acid:
HC2H3O2 (acetic acid), H2S (hydrosulfic acid). Strong acid: HNO3 (nitric acid), HCl (hydrochloric acid.)
2. Describe the difference between a weak base and a strong base. Give an example of each.

weak base partially dissociates in solution and a strong base completely dissociates in solution. Weak base:
NH3 ( ammonia), HCO3- (bicarbonate). Strong base: NaOH (sodium hydroxide), KOH (potassium hydroxide.)
3. Define the symbol Ka and Kb.

Ka is the acid ionization constant and it is the ratio of the concentrations of the products divided by the
[𝐻+][𝐴−]
reactants for an acid. If Ka>1 then it is a strong acid and < 1 a weak acid. 𝐾𝑎 = [𝐻𝐴} . Kb is the equivalent
relationship for bases.
4. The concentration H+ in a HF acid solution is 1.0 x 10-4 M and the amount of undissociated (not
ionized) Hydroflouric Acid, HF, is 2.5 x 10-6 M. What is the dissociation constant for this acid? Is this
considered to be a strong or weak acid?

+
−
𝐻𝐹(𝑔) ⟶ 𝐻(𝑎𝑞)
+ 𝐹(𝑎𝑞)
𝐾𝑎 =
[𝐻+][𝐴−]
(1 𝑥 10−4 )(1 𝑥 10−4 )
[𝐻𝐴}
2.5 𝑥10−6
=
= .004
therefore weak acid
5. Acetic Acid is the conjugate acid of the acetate anion. It is a weak monoprotic acid that dissociates to
an acetate ion and a hydrogen ion in aqueous solution. Calculate Ka for acetic acid if a 1.0 M solution
results in an equilibrium [H+] =0.0042 M.

+
𝐻𝐶2 𝐻3 𝑂2 ⟶ 𝐻(𝑎𝑞)
+ 𝐶2 𝐻3 𝑂2 −
(𝑎𝑞)
𝐾𝑎 =
[𝐻+][𝐴−]
(.0042).0042)
[𝐻𝐴}
1.0
=
= 1.76 𝑥 10−5
therefore weak acid
6. Ammonia is a weak base. If the initial concentration of ammonia is 0.150 M and the equilibrium
concentration of OH- is 1.6 x 10-3 M, calculate Kb for ammonia.

−
𝑁𝐻3 ⟶ 𝑁𝐻4 +
(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)
𝐾𝑏 =
[𝑂𝐻−][𝐻𝐵+]
(1.6 𝑥 10−3 )(1.6 𝑥 10−3 )
[𝐻𝐵]
0.150
=
= 1.71 𝑥 10−5
therefore weak
base
7. At 37°C, which is normal body temperature, Kw = 2.4 x 10-14. Calculate [H+] and [OH-] in a neutral
solution at this temperature.
 𝐾𝑤 = [𝐻 +][𝑂𝐻−]
2.4 x 10-14 = [H+][OH-]
2.4 x 10-14 = x2
= 1.55 x 10-7 M
8. At 50°C, Kw = 5.47 x 10-14. Calculate [H+] and [OH-] in a neutral solution at this temperature.
 𝐾𝑤 = [𝐻 +][𝑂𝐻−]
5.47 x 10-14 = [H+][OH-]
5.47 x 10-14 = x2
= 2.34 x 10-7 M
9. A 0.010 M solution of aspirin, a weak monoprotic acid, has a pH of 3.3. What is the Ka of aspirin?

𝑝𝐻 = − log[𝐻 + ]
3.3 = −log[𝐻 + ]
10−3.3 = 5.01 𝑥 10−4 𝑀
𝐾𝑎 =
(5.01 𝑥 10−4 )(5.01 𝑥 10−4 )
0.010
= 2.51 𝑥 10−5
10. A 0.513 M solution of a weak base has a pH of 11.4. What is the Kb of the base?

𝑝𝐻 + 𝑝𝑂𝐻 = 14
𝑝𝑂𝐻 = −log(𝑂𝐻−)
pOH=14-11.4
pOH= 2.6
2.6 = −log(𝑂𝐻 − )
10−2.6 = .0025 𝑀
𝐾𝑏 =
(.0025).0025)
0.153
= 4.08 𝑥 10−5
11. Formic acid is a weak monoprotic acid that is partly responsible for the irritation of insect bites. If the
initial concentration of formic acid is 0.1 M and the equilibrium concentration of both the H+ and the
conjugate base are 4.2 x 10-3, calculate Ka for formic acid.

𝐾𝑎 =
(4.2 𝑥 10−3 )(4.2 𝑥 10−3 )
0.1
= 1.76 𝑥 10−4
12. The acetate ion is a very weak base. If the initial concentration of acetate ion in solution is 0.1 M, and
the equilibrium concentration of OH- is 7.5 x 10-6 M, calculate Kb for the acetate ion.

𝐾𝑏 =
(7.5 𝑥 10−6 )(7.5 𝑥 10−6 )
0.1
= 5.63 𝑥 10−10
13. How many moles of LiOH are needed to exactly neutralize 2.0 moles of H2SO4?
 H2SO4(aq) + 2LiOH(aq) ⟶ 2H2O(l) + Li2SO4(aq)
2.0 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4 ×
2𝑚𝑜𝑙𝑒𝑠 𝐿𝑖𝑂𝐻
= 4.0 𝑚𝑜𝑙𝑒𝑠 𝐿𝑖𝑂𝐻
1 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4
14. How many moles of H2SO4 are needed to neutralize 5.0 moles of NaOH?
 H2SO4(aq) + 2NaOH(aq) ⟶ 2H2O(l) + Na2SO4(aq)
5.0 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 ×
1 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4
= 2.5 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
15. How many moles of HCL are needed to neutralize 0.10 L of 2.0 M NaOH?
 HCl(aq) + NaOH(aq) ⟶ H2O(l) + NaCl(aq)
2.0 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
0.10𝐿
×
×
= 0.2 𝑚𝑜𝑙 𝐻𝐶𝑙
1𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1
16. How many moles of NaOH are needed to neutralize 0.010 L of 0.20 M H2SO4?
H2SO4(aq) + 2NaOH(aq) ⟶ 2H2O(l) + Na2SO4(aq)
0.20 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4 2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0.10𝐿
×
×
= 0.04 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1𝐿
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
1
17. If it takes 15.0 mL of 0.40 M NaOH to neutralize 5.0 mL of HCl, what is the molar concentration of the
HCl solution?
 HCl(aq) + NaOH(aq) ⟶ H2O(l) + NaCl(aq)
0.40 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
0.015𝐿
1
×
×
×
= 1.2 𝑀 𝐻𝐶𝑙
1𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1
0.005𝐿
18. If it takes 10.0 mL of 2.0 M H2SO4 to neutralize 30.0 mL of KOH, what is the molar concentration of the
KOH?
 H2SO4(aq) + 2KOH(aq) ⟶ 2H2O(l) + K2SO4(aq)
2.0 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆𝑂4
2 𝑚𝑜𝑙 𝐾𝑂𝐻
0.010𝐿
1
×
×
×
= 1.33 𝑀 𝐾𝑂𝐻
1𝐿
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
1
0.030𝐿
19. How many mL of 2.0 M H2SO4 are required to neutralize 30.0 mL of 1.0 M NaOH?
 H2SO4(aq) + 2NaOH(aq) ⟶ 2H2O(l) + Na2SO4(aq)
1.0 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4 0.03𝐿
1𝐿
×
×
×
= .0075 𝐿 𝑜𝑟 7.5 𝑚𝐿
1𝐿
2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1
2.0 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
20. How many mL of 0.10 M Ca(OH)2 are required to neutralize 25.0 mL of 0.50 M HNO3?
 2HNO3(aq) +Ca(OH)2(aq) ⟶ 2H2O(l) + Ca(NO3)2(aq)
0.5 𝑚𝑜𝑙𝑒𝑠 𝐻𝑁𝑂3
1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2 0.025𝐿
1𝐿
×
×
×
= .0625 𝐿 𝑜𝑟 62.5 𝑚𝐿
1𝐿
2 𝑚𝑜𝑙 𝐻𝑁𝑂3
1
0.1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2