Honors Chemistry Ka, Kb and Neutralization Rxns. Describe the

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Honors Chemistry
Ka, Kb and Neutralization Rxns.
1. Describe the difference between a weak acid and a strong acid. Give an example of each.
ο‚·
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.
ο‚·
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.
ο‚·
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. πΎπ‘Ž = [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?
ο‚·
+
−
𝐻𝐹(𝑔) ⟢ 𝐻(π‘Žπ‘ž)
+ 𝐹(π‘Žπ‘ž)
πΎπ‘Ž =
[𝐻+][𝐴−]
(1 π‘₯ 10−4 )(1 π‘₯ 10−4 )
[𝐻𝐴}
2.5 π‘₯10−6
=
= .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.
ο‚·
+
𝐻𝐢2 𝐻3 𝑂2 ⟢ 𝐻(π‘Žπ‘ž)
+ 𝐢2 𝐻3 𝑂2 −
(π‘Žπ‘ž)
πΎπ‘Ž =
[𝐻+][𝐴−]
(.0042).0042)
[𝐻𝐴}
1.0
=
= 1.76 π‘₯ 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.
ο‚·
−
𝑁𝐻3 ⟢ 𝑁𝐻4 +
(π‘Žπ‘ž) + 𝑂𝐻(π‘Žπ‘ž)
𝐾𝑏 =
[𝑂𝐻−][𝐻𝐡+]
(1.6 π‘₯ 10−3 )(1.6 π‘₯ 10−3 )
[𝐻𝐡]
0.150
=
= 1.71 π‘₯ 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.
ο‚· 𝐾𝑀 = [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.
ο‚· 𝐾𝑀 = [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?
ο‚·
𝑝𝐻 = − log[𝐻 + ]
3.3 = −log[𝐻 + ]
10−3.3 = 5.01 π‘₯ 10−4 𝑀
πΎπ‘Ž =
(5.01 π‘₯ 10−4 )(5.01 π‘₯ 10−4 )
0.010
= 2.51 π‘₯ 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?
ο‚·
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 π‘₯ 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.
ο‚·
πΎπ‘Ž =
(4.2 π‘₯ 10−3 )(4.2 π‘₯ 10−3 )
0.1
= 1.76 π‘₯ 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.
ο‚·
𝐾𝑏 =
(7.5 π‘₯ 10−6 )(7.5 π‘₯ 10−6 )
0.1
= 5.63 π‘₯ 10−10
13. How many moles of LiOH are needed to exactly neutralize 2.0 moles of H2SO4?
ο‚· H2SO4(aq) + 2LiOH(aq) ⟢ 2H2O(l) + Li2SO4(aq)
2.0 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4 ×
2π‘šπ‘œπ‘™π‘’π‘  𝐿𝑖𝑂𝐻
= 4.0 π‘šπ‘œπ‘™π‘’π‘  𝐿𝑖𝑂𝐻
1 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4
14. How many moles of H2SO4 are needed to neutralize 5.0 moles of NaOH?
ο‚· H2SO4(aq) + 2NaOH(aq) ⟢ 2H2O(l) + Na2SO4(aq)
5.0 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Žπ‘‚π» ×
1 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4
= 2.5 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4
2 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Žπ‘‚π»
15. How many moles of HCL are needed to neutralize 0.10 L of 2.0 M NaOH?
ο‚· HCl(aq) + NaOH(aq) ⟢ H2O(l) + NaCl(aq)
2.0 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Žπ‘‚π»
1 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
0.10𝐿
×
×
= 0.2 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
1𝐿
1 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
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 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4 2 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π» 0.10𝐿
×
×
= 0.04 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
1𝐿
1 π‘šπ‘œπ‘™ 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?
ο‚· HCl(aq) + NaOH(aq) ⟢ H2O(l) + NaCl(aq)
0.40 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Žπ‘‚π»
1 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
0.015𝐿
1
×
×
×
= 1.2 𝑀 𝐻𝐢𝑙
1𝐿
1 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
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?
ο‚· H2SO4(aq) + 2KOH(aq) ⟢ 2H2O(l) + K2SO4(aq)
2.0 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑆𝑂4
2 π‘šπ‘œπ‘™ 𝐾𝑂𝐻
0.010𝐿
1
×
×
×
= 1.33 𝑀 𝐾𝑂𝐻
1𝐿
1 π‘šπ‘œπ‘™ 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?
ο‚· H2SO4(aq) + 2NaOH(aq) ⟢ 2H2O(l) + Na2SO4(aq)
1.0 π‘šπ‘œπ‘™π‘’ π‘π‘Žπ‘‚π»
1 π‘šπ‘œπ‘™ 𝐻2 𝑆𝑂4 0.03𝐿
1𝐿
×
×
×
= .0075 𝐿 π‘œπ‘Ÿ 7.5 π‘šπΏ
1𝐿
2 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
1
2.0 π‘šπ‘œπ‘™ 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?
ο‚· 2HNO3(aq) +Ca(OH)2(aq) ⟢ 2H2O(l) + Ca(NO3)2(aq)
0.5 π‘šπ‘œπ‘™π‘’π‘  𝐻𝑁𝑂3
1 π‘šπ‘œπ‘™ πΆπ‘Ž(𝑂𝐻)2 0.025𝐿
1𝐿
×
×
×
= .0625 𝐿 π‘œπ‘Ÿ 62.5 π‘šπΏ
1𝐿
2 π‘šπ‘œπ‘™ 𝐻𝑁𝑂3
1
0.1 π‘šπ‘œπ‘™ πΆπ‘Ž(𝑂𝐻)2
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