Chapter 18
Acid-Base Equilibria
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Acid – Base Reactions
• Strong acid + strong base:
• HNO3 + Ca(OH)2
•
H+(aq) + OH-(aq) ----------> H2O(l)
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Acid-Base Reactions
• Weak acid + strong base:
• HF + KOH
• HF(aq) + OH-(aq) <----> F-(aq) + H2O(l)
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Acid-Base Reactions
• Strong acid + weak base:
• HClO4 + NH3
•
H+(aq) + NH3(aq) <----------> NH4+(aq)
• H2SO4 + Na2CO3
•
H+(aq) + CO32-(aq) <------> HCO3-(aq)
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The Common Ion Effect
The common ion effect is when an ion
common to the ionization of an acid is
present in the solution in an amount
greater than that produced by the acid
ionization. The presence of this ion,
according to LeChatelier's principle, limits
the extent to which the acid will ionize and
thus affects the pH of the solution.
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Example
• Determine the pH of a solution of 0.25 M
acetic acid. Ka = 1.8 x 10-5
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Example
• Determine the pH of a solution of 0.25 M
acetic acid in a solution of 0.10 M sodium
acetate.
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Example
Determine the pH of a solution prepared by
mixing 50.0 mL of 0.100 M HOCl with 50.0 mL of
0.100 M NaOCl (Ka = 3.5x108).
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Buffers
• Acid-base buffers confer resistance to a
change in the pH of a solution when
hydrogen ions (protons) or hydroxide ions
are added or removed. An acid-base
buffer typically consists of a weak acid,
and its conjugate base.
• Prepared by adding both the weak acid
HB and the salt of its conjugate base B- to
water.
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General Expressions for Buffer
Solutions
• Assume equilibrium is established, therefore
• Ka = [H3O+] [B-] / [HB]
and
• [H3O+] = Ka [HB] /[B-]
or
• Kb = [HB] [OH-] / [B-]
and
• [OH-] = Kb [B-] /[HB]
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Example
• Determine the pH of a solution of 0.10 M
acetic acid in a solution of 0.10 M sodium
acetate.
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Effect of H3O+ or OH- on Buffer
System
• A buffer system contains one species (HB)
that will react with added hydroxide ions
and another species (B-) that will react
with added hydronium ions.
• Both reactions will go virtually to
completion hence, the added hydronium or
hydroxide ions are consumed and the
effect on the overall pH is negligible.
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Buffer Calculations
• Determine the concentrations of HB and
B- after the addition of H3O+ or OH-
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Addition of an Acid
• 0.10 M HC2H3O2 with 0.10 M NaC2H3O2
add 50.0 mL 0.10 M HCl
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Addition of a Base
• 0.10 M HC2H3O2 with 0.10 M NaC2H3O2
add 50.0 ml 0.10 M NaOH
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The Henderson-Hasselbalch
Equation
• pH = pKa + log [conjugate base]/[acid]
• This equation clearly shows that the pH of
the solution of a weak acid and its
conjugate base is controlled primarily by
the strength of the acid.
• Can also be written
• pOH = pKb + log [conjugate acid]/[base]
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Henderson-Hasselbalch
Equation
Allows us to predict pH when HB/B mixed.
• When [B] /[HB] = 1 (i.e. [HB]=[B]), pH = pKa
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Example
Calculate pH of solution containing 0.040M
Na2HPO4 and 0.080M KH2PO4. pKa2=7.20.
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Example
Determine the ratio of the concentration of the
conjugate acid to concentration of the
conjugate base for a weak acid in which the
pH was 5.45 and pKa was 5.75.
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Example
Determine the pH of a solution consisting of
0.100 M NH3 and 0.150 M NH4Cl.
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Preparation of a Buffer Solution
Selection of weak acid or weak base
• (pKa ~ pH)
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Titrations of Weak Acids and
Strong Bases
Typical net ionic equation
•
HC2H3O2 + OH- ------> C2H3O2- + H2O
•
K = 1/Kb(C2H3O2-) = 1/5.6 x 10-10 =
1.8 x 109
• The reaction goes essentially to
completion
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pH Changes
• -pH starts off at about 2.4, the pH of 1 M
HC2H3O2
• -Region, centered around the half-way point of
the titration where pH changes very slowly. In
this region there are appreciable amounts of two
different species: unreacted HC2H3O2and
C2H3O2- ions. Hence, we have a buffer system
• -Equivalence point - we have a solution of
sodium acetate. This solution is basic because
C2H3O2- is a weak base. pH at equivalence is >
7.00
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pH Calculations
• Write a net ionic equation to determine the
extent of the reaction.
• Calculate the initial pH from the Ka or Kb.
• Calculate pH at midpoint using the buffer
relation.
• Calculate pH at equivalence point using
Ka of conjugate acid or Kb of conjugate
base.
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pH Titration Curves
Titration of 0.100 M HA with
0.100 M NaOH
14
12
10
pH
• Titration curve: plot
of pH of the solution
as a function of the
volume of base (acid)
added to an acid
(base).
8
6
WA
4
SA
2
0
0
10
20
30
40
Volume Base Added, mL
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• Sharp rise in curve is equivalence
point.
• pH at equivalence point is 7.0 for SA
but higher for WA.
• Equivalence point can be used to
determine the concentration of the
titrant.
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Example
The equivalence point for 15.00 mL of
an acid occurred when 25.00 mL of
0.075 M NaOH was added. What
was the molarity of the acid?
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SA–SB Titrations
• Base removes some acid and pH increases.
• Let
nb = moles of base added
na,r = moles of acid remaining
na,r = na  nb = MaVa  MbVb
• Moles of hydronium ion same as moles of
acid remaining.
nH3O+ = na,r;
• Valid until very close to equivalence point.
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• Equivalence point(EP): pH = 7.00
• Beyond EP: pH due only to base added
(i.e. excess base). Use total volume.
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Example
• Determine pH of 10.0 mL of 0.100M HCl
after addition of 5.00, 10.0 and 15.0mL of
0.100M NaOH.
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Titration of SB with SA
• Acid removes some of the base and pH is changed by amount
of base removed.
Let na = moles of acid added
nb,r = moles of base remaining
nb,r = MbVb  MaVa
• Moles of hydroxide ion same as moles of base remaining.
• nOH = nb,r;
– Valid until EP.
• EP: pH = 7.00
• Beyond EP: pH due only to excess acid. Use total volume.
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Example
• Determine pH of 10.0 mL of 0.100M NaOH
after addition of 5.00, 10.0 and 15.0mL of
0.100M HCl.
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WA with SB Titration
• As above base removes some of the acid and pH is
changed by amount of acid removed.
Let nb = moles of base added
nHA = moles of acid remaining
nHA = MHAVHA  MbVb
nA = nb = MbVb
• Up to equivalence point moles of hydronium ions
must be determined from equilibrium expression.
• Equivalence point: pH = pH of salt of WA
• Beyond Equivalence point: Use amount of excess
base to determine pH.
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Example
• Determine pH of 10.0 mL of 0.100M HA
after addition of 5.00, 10.0 and 15.0mL of
0.100M NaOH. Ka = 1.75x105.
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WB–SA Titrations
• Acid removes some of the base and decreases the
pH.
Let na = moles of acid added
nb,r = moles of base remaining
nb,r = CbVb  CaVa
nBH+ = na = CaVa
• Moles of hydroxide ions must be determined from
equilibrium expression. Valid until EP.
• EP: pH = pH of salt of weak base.
• Beyond EP: pH due only to presence of acid added
after endpoint (i.e. excess acid) as seen for strong
base. Volume correction needed as above (total
volume).
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Example
Determine pH of 10.0 mL of 0.100M B after
addition of 5.00, 10.0 and 15.0mL of
0.100M HCl. Kb = 1.75x105.
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Acid-Base Indicators
• An acid-base indicator (HIn) is usually an
organic dye that is itself a weak acid
governed by an equilibrium constant.
• The acid form has one color and the base
form has another. In principle, the color of
the indicator changes when [H3O+] = Ka
of the indicator.(because HIn = In- at this
point)
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Choice of Indicator
1. Strong acid-Weak base
• Solution at equivalence point is weakly
acidic. Choose indicator which turns color
below pH 7. ex methyl red (pH 5)
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Choice of Indicator
2. Weak acid - Strong base
• Solution at equivalence point is weakly
basic, Choose indicator which turns color
above pH 7. ex. phenolphthalein (pH 9)
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Choice of Indicator
• Strong acid - Strong base
• Solution at equivalence point is neutral.
However, pH changes so rapidly near the
end point that any indicator that changes
color between 5 and 9 will work.
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