Chemistry 201: General Chemistry II - Lecture
Dr. Namphol Sinkaset
Chapter 17 Study Guide
Concepts
1. There are multiple definitions for acids and bases.
2. An Arrhenius acid is a substance that produces H+ (H3 O+ ) ions in aqueous solution.
An Arrhenius base is a substance that produces OH− ions in aqueous solution.
3. A Brønsted-Lowry acid is a proton donor whereas a Brønsted-Lowry base is a
proton acceptor.
4. Under the Brønsted-Lowry definitions, acids and bases always occur in pairs.
5. Substances that can act as either an acid or base are called amphoteric.
6. Conjugate acid-base pairs are two substances related to each other by a transfer of
a proton.
7. A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.
8. A strong acid completely ionizes in solution.
9. A weak acid partially ionizes in solution.
10. Strong acids include: hydrochloric, hydrobromic, hydroiodic, nitric, chloric, perchloric,
bromic, perbromic, iodic, periodic, and sulfuric acids.
11. Weak acids establish an equilibrium between their ionized and nonionized forms.
12. The stronger the acid, the weaker its conjugate base.
13. An acid dissociation constant, Ka , is an equilibrium constant written for the equilibrium set up by a weak acid.
14. Water reacts with itself in an autoionization reaction.
15. The ion product constant for water, Kw , is the equilibrium constant written for
the autoionization reaction of water. Note that Kw = 1.00 × 10−14 .
16. When [H3 O+ ] = [OH− ], a solution is neutral.
17. Acidic solutions have additional H3 O+ ions. Basic solutions have additional OH−
ions. Regardless, the product of hydronium and hydroxide ions always equals Kw .
18. pH is another way to specify the acidity of a solution. pH < 7 is acidic, pH = 7 is
neutral, and pH > 7 is basic.
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19. The log function has its own significant figure rule.
20. The p function is the mathematical operation of taking the negative log.
21. Weak acid equilibrium problems can be solved using techniques applied to ordinary
equilibria. In these problems, we ignore the insignificant contribution of H3 O+ from
the autoionization of water.
22. The strength of a weak acid can also be characterized by its percent ionization.
23. A mixture of a strong acid with one or more weak acids can be treated as if the strong
acid is the only component.
24. A mixture of weak acids may need to be solved as a double equilibrium problem
depending on their relative Ka ’s.
25. Everything discussed about acids applies to bases.
26. Strong bases ionize completely. Most hydroxides of Group 1 and Group 2 metals are
strong bases.
27. Weak bases produce OH− by pulling a proton off water.
28. The strength of a weak base depends on its base ionization constant, Kb . It is the
equilibrium constant for a weak base.
29. Some ions can act as either weak acids or weak bases.
30. Anions that are conjugate bases of weak acids are themselves weak bases.
31. Anions that are conjugate bases of strong acids are pH neutral.
32. Conjugate acid/base pairs have related Ka ’s and Kb ’s.
33. Cations that are conjugate acids of weak bases are themselves weak acids.
34. Small, highly-charged metal cations form weakly acidic solutions.
35. Salt solutions can be either neutral, acidic, or basic. If the cation is a weak acid, and
the anion is a weak base, then the pH depends on the Ka and Kb .
36. Acids that have more than one acidic proton are called polyprotic acids.
37. Generally, successive Ka ’s are much smaller than the 1st, so polyprotic acids only need
to be solved for the first ionization. Exceptions include sulfuric acid and when Ka ’s are
within a few hundred of each other. Exceptions must be solved as a double equilibrium
problem.
38. The strength of a binary acid depends on bond polarity and bond strength. When
H holds the partial negative charge and the bond is relatively weak, the acid will be
strong.
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39. Oxyacid (oxoacid) strength depends on the electronegativity of Y and the number of
O atoms attached to Y. Oxyacids that have Y’s with higher electronegativities and
more O atoms are stronger acids.
Equations
1. Ka =
[H3 O+ ][A− ]
(General acid dissociation expression)
[HA]
2. Kw = [H3 O+ ][OH− ] (Ion product constant for water)
3. pH = − log [H3 O+ ] (Definition of pH)
4. pH + pOH = 14.00 (Relationship between pH and pOH)
5. % ionization =
6. Kb =
[H O+ ]equil
[ionized acid]
× 100% (% ionization)
× 100% = 3
[initial acid]
[HA]init
[BH+ ][OH− ]
(General base association equation)
[B]
7. Ka Kb = Kw = 1.00 × 10−14 (Relationship between Ka and Kb )
8. pKa = − log Ka (pKa expression)
9. pKb = − log Kb (pKb expression)
Representative Problems
R43. What are the concentrations of all the substances in a 1.0 M solution of
hydrogen peroxide, H2 O2 ? What is the pH of the solution? For H2 O2 , Ka = 1.8
× 10−12 .
We determine that H2 O2 must be a weak acid because they give us the Ka . The acid
dissociation equation is then:
−
H2 O2 (aq) + H2 O(l) *
) H3 O+
(aq) + HOO(aq)
The acid dissociation constant is then:
Ka =
[H3 O+ ][HOO− ]
[H2 O2 ]
We set up our chart and fill it in as usual.
Reaction: H2 O2(aq)
Initial
1.0
Change
−x
Equil. 1.0 − x
+
H2 O(l)
—
—
—
3
*
) H3 O+
(aq)
0
+x
x
+
HOO−
(aq)
0
+x
x
We put the equilibrium concentrations into the Ka expression.
[H3 O+ ][HOO− ]
[H2 O2 ]
(x)(x)
1.8 × 10−12 =
1.0 − x
x2
1.8 × 10−12 =
1.0
x = 1.34 × 10−6
Ka =
Checking our approximation, we get 0.00013%, well within the 5% requirement. The
concentrations of everything in solution are: [H+ ] = 1.3 × 10−6 M, [HOO− ] = 1.3 × 10−6
M, [H2 O2 ] = 1.0 M. (You’d have to go out to 0.9999987 M to see a difference, but of course,
sig figs limit us from going out that far).
To find the pH, we just take the -log of the [H+ ].
pH =
=
=
=
− log [H+ ]
− log(1.34 × 10−6 )
5.873
5.87
R55. Quinine, an important drug in treating malaria, is a weak Brønsted base
that we may represent as Qu. To make it more soluble in water, it is put into a
solution as its conjugate acid, which we may represent as H-QuCl. What is the
calculated pH of a 0.15 M solution of H-Qu+ ? Its pKa is 8.52 at 25 ◦ C.
Another equilibrium problem, and we use the exact same steps we’ve been using. First,
identify the equilibrium reaction and write its Ka expression.
+
*
H-Qu+
(aq) + H2 O(l) ) H3 O(aq) + Qu(aq)
Ka =
[H3 O+ ][Qu]
[H-Qu+ ]
We set up the chart, filling it in with the given information.
Reaction: H-Qu+
(aq) +
Initial
0.15
Change
−x
Equil.
0.15 − x
H2 O(l)
—
—
—
4
*
) H3 O+
(aq)
0
+x
x
+ Qu−
(aq)
0
+x
x
Before we fill in everything into the Ka expression, we need to get the value of Ka from
the given pKa .
pKa
8.52
−8.52
Ka
=
=
=
=
− log Ka
− log Ka
log Ka
3.02 × 10−9
Ka =
3.02 × 10−9 =
3.02 × 10−9 =
x =
[H+ ][Qu]
[H-Qu+ ]
(x)(x)
0.15 − x
x2
0.15
2.13 × 10−5
Checking our approximation, we get 0.014%. Now, we calculate the solution pH.
pH =
=
=
=
− log [H+ ]
− log(2.13 × 10−5 )
4.671
4.67
R102. What is [OH− ] in an aqueous solution that is 5.0% NaClO by mass and
also 0.0050% NaOH by mass? What is the pH of the solution? (Assume d of
solution = 1.0 g/mL.)
In water, NaClO dissociates into Na+ and ClO− ions. Since Na+ cannot donate protons
and isn’t highly-charged, it is not a weak acid and thus is a neutral cation. ClO− can accept
a proton to form HClO, hypochlorous acid. This means that ClO− is a weak base. We write
a base equilibrium for the ClO− in water.
−
*
ClO−
(aq) + H2 O(l) ) HClO(aq) + OH(aq)
Thus, we need to calculate the concentration of ClO− .
1.0 g solution
1 mole NaClO
1000 mL solution
5.0 g NaClO
×
×
×
= 0.672 M
100.0 g solution 1.0 mL solution 74.44 g NaClO
1 L solution
Since NaClO dissociates completely, [ClO− ] = 0.0672 M. Additionally, there is NaOH
in the solution. Sodium hydroxide is a strong base, so it dissociates completely into Na+
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(which again is a neutral cation) and OH− . Since OH− appears in the equilibrium, we need
to know its concentration.
0.0050 g NaOH
1.0 g solution
1 mole NaOH
1000 mL solution
×
×
×
= 0.00125 M
100.0 g solution 1.0 mL solution 39.998 g NaOH
1 L solution
Again, since NaOH dissociates completely, [OH− ] = 0.001250 M. Now we set up our
equilibrium chart and fill in.
Reaction:
Initial
Change
Equil.
OCl−
(aq) +
0.672
−x
0.672 − x
H2 O(l)
—
—
—
*
) HClO(aq)
0
+x
x
+
OH−
(aq)
0.00125
+x
x + 0.00125
To calculate, we need the Kb . Conjugate bases are rarely found in a Kb table, so we need
to use the Ka of HClO to find the Kb of OCl− . Consulting an acid dissociation constant
table, we find that Ka = 2.9 × 10−8 for HClO.
Ka × Kb = Kw
(2.9 × 10−8 )Kb = 1.00 × 10−14
Kb = 3.45 × 10−7
Now we set up our Kb expression and solve.
[HClO][OH− ]
[OCl− ]
(x)(x + 0.00125 )
3.45 × 10−7 =
0.672 − x
0.0012
5x
3.45 × 10−7 =
0.672
x = 1.85 × 10−4
Kb =
Before continuing, we verify that our approximation is valid. Comparing x to 0.00125 ,
we find that it is 14.8 %, so we’re in trouble. We have to go through and use the quadratic
equation.
(x)(x + 0.00125 )
0.672 − x
−7
2
(3.45 × 10 )(0.0672 − x) = x + 0.00125 x
2.32 × 10−8 − (3.45 × 10−7 )x = x2 + 0.00125 x
0 = x2 + 0.00125 x − 2.32 × 10−8
3.45 × 10−7 =
Now we solve for x with the quadratic equation.
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x =
=
√
b2 − 4ac
2a q
−0.00125 ± (0.00125)2 − 4(1)(−2.32 × 10−8 )
−b ±
2(1)
−0.00125 ± 0.00129
2
= 2.00 × 10−4 or -1.27 × 10−3
=
The negative root doesn’t make physical sense, so x = 2.00 × 10−4 (our sig figs are limited
by what’s in Kb ). From our chart, [OH− ] = 1.45 ×10−3 M which equals 1.5×10−3 M reported
to the correct number of sig figs. We can calculate pH by going through pOH.
pOH = − log [OH− ]
= − log(1.45 × 10−3 )
= 2.839
Of course, pH + pOH = 14.00, so pH of the solution is 11.16.
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