Jeffrey Mack California State University, Sacramento Chapter 18
Principles of Chemical Reactivity: Other Aspects of Aqueous
Equilibria
Slide 2
More About Chemical Equilibria: Acid Base & Precipitation
Reactions
Slide 3
Stomach Acidity & AcidBase Reactions
Slide 4
In the previous chapter, you examined the behavior of weak
acids and bases in terms of equilibrium involving conjugate pairs.
The pH of a solution was found via K a or K b. What would happen if
you started with a solution of acid that was mixed with a solution
of its conjugate base? The change of pH when a significant ammout
of conjugate base is present is an example of the Common Ion
Effect. The Common Ion Effect
Slide 5
What is the effect on the pH of a 0.25M NH 3 (aq) solution when
NH 4 Cl is added? NH 4 + is an ion that is COMMON to the
equilibrium. Le Chatelier predicts that the equilibrium will shift
to the left to reduce the disturbance. This results in a reduciton
of the hydroxide ion concentration, which will lower the pH. Hint:
NH 4 + is an acid! The Common Ion Effect
Slide 6
First lets find the pH of a 0.25M NH 3 (aq) Solution: [NH 3 ]
[NH 4 + ] [OH - ] Initial0.2500 Change x +x Equilibrium 0.025 x xx
The Common Ion Effect
Slide 7
First lets find the pH of a 0.25M NH 3 (aq) Solution: [NH 3 ]
[NH 4 + ] [OH - ] Initial0.2500 Change x +x Equilibrium 0.025 x xx
The Common Ion Effect
Slide 8
First lets find the pH of a 0.25M NH 3 (aq) Solution: Assuming
x is
Consider the acetic acid / acetate buffer system. The ability
for the acid to consume OH is seen from the reverse of the base
hydrolysis: K rev is >> 1, indicating that the reaction is
product favored. An hydroxide added will immediately react with the
acid so long as it is present. Controlling pH: Buffer
Solutions
Slide 16
Consider the acetic acid / acetate buffer system. Similarly,
the conjugate base (acetate) is readily capable of consuming H 3 O
+ K rev is >> 1, indicating that the reaction is product
favored. An hydronium ion added will immediately react wit the acid
so long as it is present. Controlling pH: Buffer Solutions
Slide 17
Problem: What is the pH of a buffer that has [CH 3 CO 2 H] =
0.700 M and [CH 3 CO 2 ] = 0.600 M? Buffer Solutions
Slide 18
Problem: What is the pH of a buffer that has [CH 3 CO 2 H] =
0.700 M and [CH 3 CO 2 ] = 0.600 M? Since the concentration of acid
is greater than the base, equilibrium will move the reaction to the
right. Buffer Solutions
Slide 19
Problem: What is the pH of a buffer that has [CH 3 CO 2 H] =
0.700 M and [CH 3 CO 2 ] = 0.600 M? Assuming that x
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution?Solution: At the equivalence point, all of
the HBz is converted to Bz - by the strong base. The conjugate base
of a weak acid (Bz - ) will hydrolyze to reform the weak acid (K b
). The pH will be > 7 This will yield the [H 3 O + ] and
pH.
Slide 56
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? Volume of OH - added to the eq. point: HBz
(aq) + OH (aq) Bz (aq) + H 2 O(l) HBz (aq) + OH (aq) Bz (aq) + H 2
O(l) The new total volume of the solution is 125 mL Titration of a
Weak Acid with a Strong Base
Slide 57
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? Moles of OH - & Bz - at the eq. point:
HBz (aq) + OH (aq) Bz (aq) + H 2 O(l) The concentration of Bz - at
the eq. point is: Titration of a Weak Acid with a Strong Base
Slide 58
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? [OH - ] at the eq. point: [Bz - ][HBz][OH -
] Initial0.02000 Change- x+ x Equilibrium0.020 - xxx Titration of a
Weak Acid with a Strong Base
Slide 59
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? [OH - ] at the eq. point: [Bz - ][HBz][OH -
] Initial0.02000 Change- x+ x Equilibrium0.020 - xxx Titration of a
Weak Acid with a Strong Base
Slide 60
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? [OH - ] at the eq. point: Assuming that
x
Conclusion: Conclusion: At the equivalence point of the
titration, unlike the titration of a strong acid and strong base,
the pH is > 7. This is due to the production of the conjugate
base of a week acid. Equivalence point pH = 8.25 Equivalence point
pH = 8.25 Titration of a Weak Acid with a Strong Base
Slide 62
Conclusion: Conclusion: What would the pH equal at the half-way
point of the titration? Hint: Only of the moles of weak acid have
been converted to its conjugate base! Half-way point pH = ??
Half-way point pH = ?? Titration of a Weak Acid with a Strong
Base
Slide 63
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? At the half-way point, moles of Hbz and Bz -
are equal. This is a BUFFER SOLUTION!
Slide 64
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? At the half-way point, moles of Hbz and Bz -
are equal. This is a BUFFER SOLUTION! Titration of a Weak Acid with
a Strong Base
Slide 65
Problem: 100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution? At the half-way point, moles of Hbz and Bz -
are equal. This is a BUFFER SOLUTION! Titration of a Weak Acid with
a Strong Base
Slide 66
Acetic Acid Titrated with NaOH
Slide 67
n equivalence points In the case of a titration of a weak
polyprotic acid (H n A) there are n equivalence points. two In the
case of the diprotic oxalic acid, (H 2 C 2 O 4 ) there are two
equivalence points. Titration of a Weak Polyprotic Acid with a
Strong Base
Slide 68
The titration of a polyprotic weak acid follows the same
process a monoprotic weak acid. As the acid is titrated, buffering
occurs until the last eq. point is reached. The pH is relative to
the amounts of conjugate acids and bases. At the second eq. point
all of the acid has been converted to A 2-, pH is determined by K
b. Titration of a Weak Polyprotic Acid with a Strong Base
Slide 69
In the case of a titration of a weak base, the process follows
that of a weak acid in reverse. There exists a region of buffering
followed by a rapid drop in pH at the eq. point. Titration of a
Weak Base with a Strong Acid
Slide 70
pH Indicators for AcidBase Titrations
Slide 71
An acid/base indicator is a substance that changes color at a
specific pH. HInd (acid) has another color than Ind (base) These
are usually organic compounds that have conjugated pi-bonds, often
they are dyes or compounds that occur in nature such as red cabbage
pigment or tannins in tea. end point of the titrationCare must be
taken when choosing an appropriate indicator so that the color
change (end point of the titration) is close to the steep portion
of the titration curve where the equivalence point is found. pH
Indicators for AcidBase Titrations
Slide 72
Slide 73
Neutral pH pH7 Natural Indicators: Red Rose Extract in
Methanol
Slide 74
Prior to this chapter, exchange reactions which formed ionic
salts were governed by the solubility rules. A compound was either
soluble, insoluble or slightly soluble. So how do we differentiate
between these? The answer lies in equilibrium. It turns out that
equilibrium governs the solubility of inorganic salts. Solubility
of Salts Lead(II) iodide
Slide 75
K sp The extent of solubility can be measured by the
equilibrium process of the salts ion concentrations in solution, K
sp. K sp is called the solubility constant for an ionic compound.
It is the product of the ions solubilities. For the salt: A x B y
(s) xA y+ (aq) + yB x- (aq) K sp = [A y+ ] x [B x- ] y Solubility
of Salts
Slide 76
Slide 77
Consider the solubility of a salt MX: If MX is added to water
then: Generally speaking, If K sp >> 1 then MX is considered
to be soluble If K sp