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Acid-Base Reactions
Conjugates
do not react!!
Part 1
2
Stomach Acidity &
Acid-Base Reactions
ACIDS-BASE REACTIONS
• For any acid-base reaction where only one hydrogen ion
is transferred, the equilibrium constant for the reaction
can be calculated and is often called Knet.
• For the general reaction:
HA + B
K net
+
-
HB + A
K aHa
K aHa  K bB
=
=
K aHB+
Kw
• Knet is always Ka (reactant acid) / Ka (product acid)
• When Knet >> 1, products are favored.
• When Knet << 1, reactants are favored.
See 16.5 for manipulating K
3
•
ACIDS-BASE REACTIONS
There are four classifications or types of
4
reactions: strong acid with strong base,
strong acid with weak base, weak acid with
strong base, and weak acid with weak base.
• NOTE: For all four reaction types the
limiting reactant problem is carried out
first. Once this is accomplished, one must
determine which reactants and products
remain and write an appropriate
equilibrium equation for the remaining
mixture.
5
STRONG ACID WITH STRONG BASE
The net reaction is:
H
+
K net
+ OH  H 2O
-
1
14
=
= 1.0 10
Kw
The product, water, is neutral.
6
STRONG ACID WITH WEAK BASE
The net reaction is:
H3O + B  HB + H 2O
+
K net =
1
K aHB+
+
K bB
=
, with 1 < K net < 1.0 1014
Kw
The product is HB+ and the solution is acidic.
WEAK ACID WITH STRONG BASE
7
The net reaction is:
HA + OH
K net
-
 H 2O + A
-
K aHA
1
=
=
, with 1 < K net < 1.0 1014
KW
K bA-
The product is A- and the solution is basic.
8
WEAK ACID WITH WEAK
BASE
The net reaction is:
HA + B
K net
+
-
HB + A
K aHA
14
=
, with 1 < K net < 1.0 10
K aHB+
Notice that Knet may even be less than one.
This will occur when Ka HB+ > Ka HA.
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
to the equivalence point (mol HBz = mol
NaOH). What is the pH of the final solution?
Note: HBz and NaOH are used up!
HBz + NaOH ---> Na+ + Bz- + H2O
Ka = 6.3 × 10-5
Kb = 1.6 × 10-10
C6H5CO2H = HBz
Benzoate ion = Bz-
9
Acid-Base Reactions
The product of the titration of benzoic
acid, the benzoate ion, Bz-, is the
conjugate base of a weak acid.
The final solution is basic.
Bz- + H2O
HBz + OH-
+
+
Kb = 1.6 x 10-10
10
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
to the equivalence point. What is the pH of
the final solution?
Strategy — find the concentration of the
conjugate base Bz- in the solution AFTER
the titration, then calculate pH.
This is a two-step problem:
1st.
stoichiometry of acid-base reaction
2nd. equilibrium calculation
11
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
to the equivalence point. What is the pH of
the final solution?
STOICHIOMETRY PORTION
1. Calculate moles of NaOH required.
(0.100 L HBz)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2. Calculate volume of NaOH required.
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH required
12
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
STOICHIOMETRY PORTION
25 mL of NaOH required
3. Moles of Bz- produced = moles HBz =
0.0025 mol Bz4. Calculate concentration of Bz-.
There are 0.0025 mol of Bz- in a
TOTAL SOLUTION VOLUME of 125 mL
[Bz-] = 0.0025 mol / 0.125 L = 0.020 M
13
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
to the equivalence point (mol HBz = mol
NaOH). What is the pH of the final solution?
Note: HBz and NaOH are used up!
HBz + NaOH ---> Na+ + Bz- + H2O
Ka = 6.3 × 10-5
Kb = 1.6 × 10-10
C6H5CO2H = HBz
Benzoate ion = Bz-
14
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OH-
Kb = 1.6 x 10-
10
initial
change
equilib
[Bz-]
[HBz]
[OH-]
0.020
0
0
+x
+x
x
x
-x
0.020 - x
15
Acid-Base Reactions
16
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
EQUILIBRIUM PORTION
Bz- + H2O
equilib
HBz + OH[Bz-]
[HBz]
0.020 - x
x
K b = 1.6 x 10
-10
=
Kb = 1.6 x 10-10
[OH-]
x
x
2
0.020 - x
Solving in the usual way, we find
x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
What is the pH at the half-way point?
HBz + H2O
H3O+ + Bz-
Ka = 6.3 x 10-5
[H3O+] = { [HBz] / [Bz-] } Ka
At the half-way point, [HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20
17
The Common Ion Effect
QUESTION: What is the effect on the pH
of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
Here we are adding an ion COMMON to
the equilibrium.
Le Chatelier predicts that the equilibrium
will shift to the ____________.
LEFT
Closer to 0
The pH will go _______.
After all, NH4+ is an acid!
18
The Common Ion Effect
QUESTION: What is the effect on the
pH of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
Let us first calculate the pH of a
0.25 M NH3 solution.
[NH3]
[NH4+]
[OH-]
initial
0.25
0
0
change
-x
+x
+x
equilib
0.25 - x
x
x
19
The Common Ion Effect
QUESTION: What is the effect on the pH
of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
2
+ ][OH- ]
[NH
x
4
Kb = 1.8 x 10-5 =
=
[NH3 ]
0.25 - x
Assuming x is << 0.25, we have
[OH-] = x = [Kb(0.25)]1/2 = 0.0021 M
This gives pOH = 2.67 and so - - pH = 14.00 - 2.67 =
11.33 for 0.25 M NH3
20
The Common Ion Effect
QUESTION: What is the effect on the pH
of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
We expect that the pH will decline on
adding NH4Cl. Let’s test that at 0.10M !
[NH3]
[NH4+]
[OH-]
initial
change
equilib
0.25
-x
0.25 - x
0.10
+x
0.10 + x
0
+x
x
21
The Common Ion Effect
QUESTION: What is the effect on the pH
of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
+ ][OH- ]
[NH
x(0.10 + x)
4
Kb = 1.8 x 10-5 =
=
[NH3 ]
0.25 - x
Because equilibrium shifts left, x is MUCH
less than 0.0021 M, the value without NH4Cl.
[OH-] = x = (0.25 / 0.10)Kb = 4.5 x 10-5 M
This gives pOH = 4.35 and pH = 9.65
pH drops from 11.33 to 9.65 on
adding a common ion.
22
Buffer Solutions
HCl is added to
pure water.
HCl is added to a
solution of a weak
acid H2PO4- and its
conjugate base
HPO42-.
23
Buffer Solutions
The function of a buffer is to resist changes
in the pH of a solution.
Buffers are just a special case of the
common ion effect.
Buffer Composition
Weak Acid
+
Conj. Base
HC2H3O2
+
C2H3O2H2PO4+
HPO42Weak Base
+
Conj. Acid
NH3
+
NH4+
24
Buffer Solutions
Consider HOAc/OAc- to see how buffers work.
ACID USES UP ADDED OH .
We know that
OAc- + H2O
HOAc + OH
has Kb = 5.6 x 10-10
Therefore, the reverse reaction of the
WEAK ACID with added OH- has
Kreverse = 1/ Kb = 1.8 x 109
Kreverse is VERY LARGE, so HOAc completely
uses up the OH !!!!
25
26
Buffer Solutions
Consider HOAc/OAc- to see how buffers work.
+
CONJUGATE BASE USES UP ADDED H
+
HOAc + H2O
OAc + H3O
has Ka = 1.8 x 10-5.
Therefore, the reverse reaction of the
+
WEAK BASE with added H has
Kreverse = 1/ Ka = 5.6 x 104
Kreverse is VERY LARGE, so OAc- completely
+
uses up the H !
Buffer Solutions
27
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O
OAc- + H3O+
Ka = 1.8 x 10-5
initial
change
equilib
[HOAc]
[OAc-]
[H3O+]
0.700
0.600
0
-x
0.700 - x
+x
0.600 + x
+x
x
Buffer Solutions
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O
OAc- + H3O+
Ka = 1.8 x 10-5
[HOAc]
[OAc-]
[H3O+]
equilib
0.700 - x
0.600 + x
x
Assuming that x << 0.700 and 0.600, we have
+ ](0.600)
[H
O
3
Ka = 1.8 x 10-5 =
0.700
[H3O+] = 2.1 x 10-5 and pH = 4.68
28
Buffer Solutions
Notice that the expression for calculating the
H+ concentration of the buffer is
+
[H3O ] =
Orig. conc. of HOAc
Orig. conc. of OAc
• Ka
This leads to a general equation for finding the
H+ or OH- concentration of a buffer.
+
[Acid]
[H3O ] =
• Ka
[Conj. base]
[OH- ] =
[Base]
• Kb
[Conj. acid]
Notice that the H+ or OH- concentrations
depend on K and the ratio of acid and
base concentrations.
29
Henderson-Hasselbalch
Equation
[Acid]
[H3O ] =
• Ka
[Conj. base]
+
Take the negative log of both sides of this
equation
[Acid]
pH = pKa - log
[Conj. base]
OR
[Conj. base]
pH = pKa + log
[Acid]
This is called the Henderson-Hasselbalch
equation.
30
31
Henderson-Hasselbalch
Equation
[Conj. base]
pH = pK a + log
[Acid]
This shows that the pH is
determined largely by the pKa of
the acid and then adjusted by the
ratio of acid and conjugate base.
Adding an Acid to a Buffer
Problem: What is the new pH when
1.00 mL of 1.00 M HCl is added to:
a) 1.00 L of pure water (before HCl, pH = 7.00)
b) 1.00 L of buffer that has [HOAc] = 0.700 M
and [OAc-] = 0.600 M (pH = 4.68)
Solution to Part (a)
Calculate [HCl] after adding 1.00 mL of HCl to
1.00 L of water
M1 • V1 = M2 • V2
M2 = 1.00 x 10-3 M
pH = 3.00
32
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b)
Step 1 — do the stoichiometry
H3O+ (from HCl) + OAc- (from buffer) --->
HOAc (from buffer)
The reaction occurs completely because K is
very large.
33
Adding an Acid to a Buffer
34
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 1—Stoichiometry
[H3O+]
[OAc-]
[HOAc]
0.00100
0.600
0.700
Change
-0.00100
-0.00100
After rxn
0
0.599
Before rxn
+0.00100
0.701
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to;
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O
H3O+ + OAc[HOAc]
[OAc-]
0.701
0.599
0
Change
-x
+x
+x
After rxn
0.710-x
Before rxn
0.599+x
[H3O+]
x
35
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O
H3O+ + OAc[HOAc]
[OAc-]
[H3O+]
After rxn
0.710-x
0.599+x
x
Because [H3O+] = 2.1 x 10-5 M BEFORE
adding HCl, we again neglect x relative to
0.701 and 0.599.
36
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O
OAc- + H3O+
+
[HOAc]
0.701
[H3O ] =
• Ka =
• (1.8 x 10 -5 )
0.599
[OAc - ]
[H3O+] = 2.1 x 10-5 M ------> pH = 4.68
The pH has not changed significantly upon
adding HCl to the buffer!
37
Preparing a Buffer
38
You want to buffer a solution at pH = 4.30.
This means [H3O+] = 10-pH = 5.0 x 10-5 M
It is best to choose an acid such that
[H3O+] is about equal to Ka (or pH pKa).
You get the exact [H3O+] by adjusting the
ratio of acid to conjugate base.
39
Preparing a Buffer Solution
Buffer prepared from
HCO3
weak acid
CO32conjugate base
-
HCO3 + H2O
+
H3O + CO32-
40
Preparing a Buffer
You want to buffer a solution at
pH = 4.30 or [H3O+] = 5.0 x 10-5 M
POSSIBLE ACIDS
Ka
HSO4- / SO42-
1.2 x 10-2
HOAc / OAc-
1.8 x 10-5
HCN / CN-
4.0 x 10-10
Best choice is acetic acid / acetate.
41
Preparing a Buffer
You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M
+
[HOAc]
-5
[H3O ] = 5.0 x 10
=
(1.8 x 10 -5 )
[OAc - ]
Solve for [HOAc]/[OAc-] ratio = 2.78/ 1
Therefore, if you use 0.100 mol of NaOAc
and 0.278 mol of HOAc, you will have
pH = 4.30.
42
Preparing a Buffer
A final point —
CONCENTRATION of the acid and
conjugate base are not important.
It is the RATIO OF THE NUMBER OF
MOLES of each.
This simplifying approximation will be
correct for all buffers with 3<pH<11,
since the [H]+ will be small compared
to the acid and conjugate base.
REVIEW PROBLEMS
• Calculate the pH of a 0.10 M HNO2
solution before and after making the
solution 0.25 M in NaNO2.
• Calculate the pH of 0.500 L of a buffer
solution composed of 0.10 M sodium
acetate and 0.15 M acetic acid, before
and after adding 1.0 grams of sodium
hydroxide solid (no volume change).
43
REVIEW PROBLEMS
• Calculate the pH of a solution that is
0.18 M in Na2HPO4 and 0.12 M in
NaH2PO4.
6.2 x 10-8
• Suggest an appropriate buffer system
for pH 5.0.
SOLUTIONS
44
Titrations
pH
Titrant volume, mL
45
Acid-Base Titrations
Adding NaOH from the buret to acetic acid in
the flask, a weak acid.
In the beginning the pH increases very slowly.
46
47
Acid-Base Titrations
Additional NaOH is added.
pH rises as equivalence point is
approached.
48
Acid-Base Titrations
Additional NaOH is added.
pH increases and then levels off as NaOH
is added beyond the equivalence point.
49
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Equivalence
point
pH of solution of
benzoic acid, a
weak acid
Acid-Base Titrations
50
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence
point. What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OHKb = 1.6 x 10-10
equilib
[Bz-]
0.020 - x
[HBz]
x
[OH-]
x
2
x
Kb = 1.6 x 10-10 =
0.020 - x
Solving in the usual way, we find:
x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25
51
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Half-way
point
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
What is the pH at the half-way point?
HBz + H2O
H3O+ + Bz-
Ka = 6.3 x 10-5
[H3O+] = { [HBz] / [Bz-] } Ka
At the half-way point, [HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20
52
Sample Problem
53
Titration Curve
20.00 mL 0.30 M HC2H3O2 is titrated
with 0.30 M NaOH. Calculate the pH at
0 mL NaOH added
HC2H3O2 <---> H+ + C2H3O20.30
0
0
-x
+x
+x
0.30
x
x
Ka =
[C2H3O2-][H+]
[HC2H3O2]
x2
=
X = 2.3 x 10-3 M = [H+]
(0.30)
= 1.8 x 10-5
pH = 2.63
Sample Problem
5.0 mL NaOH added
HC2H3O2- + OH- ---> C2H3O2- + HOH
6.0
1.5
0
-1.5
4.5
-1.5
0
+ 1.5
1.5
[C2H3O2- ] =
1.5
= 0.060 M
25.00
[HC2H3O2 ] =
4.5
25.00
= 0.18 M
54
Sample Problem
HC2H3O2
Ka =
<--->
H+
+
C2H3O2-
0.18
0
0.060
-x
0.18
+x
x
+x
0.060
[C2H3O2-][H+]
[HC2H3O2]
=
X = 5.4 x 10-5 M = [H+]
55
x(0.060)
(0.18)
= 1.8 x 10-5
pH = 4.27
Sample Problem
20.0 mL NaOH added
HC2H3O2 + OH- ---> C2H3O2- + HOH
6.0
6.0
0
-6.0
0
-6.0
0
+ 6.0
6.0
[C2H3O2- ] =
6.0
40.00
= 0.15 M
56
Sample Problem
57
C2H3O2- + HOH <---> OH- + HC2H3O2
0.15
0
0
-x
0.15
+x
x
+x
x
Kb =
[C2H3O2-][H+]
[HC2H3O2]
=
X = 9.2 x 10-6 M = [OH-]
x2
(0.15)
= 5.6 x 10-10
pH = 8.96
Sample Problem
30.0 mL NaOH added
HC2H3O2 + OH- ---> C2H3O2- + HOH
6.0
9.0
0
-6.0
0
-6.0
3.0
+ 6.0
6.0
[OH-] =
3.0
= 0.060 M
50.0
pOH = 1.22
pH = 12.78
58
Sample Problem
Titration Curve
25.00 mL 0.300 M HCl is titrated with
0.7500 M NaOH. Calculate the pH at:
0 mL NaOH added
[H+] = 0.300 M pH = 0.523
59
Sample Problem
5.00 mL NaOH added
HCl +
NaOH
12.50
3.75
-3.75
8.75
-3.75
0
[H+]
=
8.75
---> NaCl + HOH
= 0.292 M
30.00
pH = 0.535
60
Sample Problem
16.67 mL NaOH added
HCl +
NaOH
12.50
12.50
-12.50
0
-12.50
0
---> NaCl + HOH
pH = 7.00
61
Sample Problem
20.00 mL NaOH added
HCl +
NaOH
12.50
15.00
-12.50
0
-12.50
2.50
[OH-]
=
2.50
---> NaCl + HOH
= 0.0556 M
45.00
pH = 12.745
62
Sample Problem
63
Titration Curve
20.00 mL 0.150 M NH3 is titrated
with 0.100 M HCl. Calculate the pH at:
0 mL HCl added
NH3 + HOH <---> NH4+ + OH0.150
0
0
-x
+x
+x
0.150
x
x
Kb =
[NH4+][OH-]
[NH3]
x2
=
X = 1.6 x 10-3 M = [OH-]
(0.150)
= 1.8 x 10-5
pH = 11.20
Sample Problem
10.0 mL HCl added
NH3
+
H+ ---> NH4+
3.00
1.00
0
-1.00
2.00
-1.00
0
+ 1.00
1.00
[NH4+] =
[NH3] =
1.00
= 0.0333 M
30.0
2.00
30.0
= 0.0667 M
64
Sample Problem
NH4+
<--->
0.0333
-x
0.0333
Ka =
NH3
+
[NH4+]
H+
0.0667
+x
0.0667
[NH3] [H+]
65
=
+x
x
(0.0667)x
X = 2.8 x 10-10 M = [H+]
(0.0333)
= 5.6 x 10-10
pH = 9.55
Sample Problem
30.0 mL HCl added
NH3
+
H+ ---> NH4+
3.00
3.00
0
-3.00
0
-3.00
0
+ 3.00
3.00
[NH4+] =
3.00
50.00
= 0.0600 M
66
Sample Problem
NH4+
<--->
0.0600
-x
0.0600
Ka =
[NH3] [H+]
[NH4+]
NH3
H+
+
0
0
+x
+x
x
x
=
X = 5.8 x 10-6 M = [H+]
67
x2
(0.0600)
= 5.6 x 10-10
pH = 5.24
Sample Problem
40.0 mL HCl added
NH3
+
H+ ---> NH4+
3.00
4.00
0
-3.00
-3.00
+ 3.00
0
1.00
3.00
[H+] =
1.00
60.00
= 0.0167
pH = 1.78
68
69
Practice Problems
1. Determine the pH of a solution made by
mixing 25.0 mL of 0.20 M nitric acid with
25.0 mL of 0.10 M potassium hydroxide.
2. Determine the pH at the equivalence point if
20.0 mL of 0.30 M HCN is titrated with 0.20 M
sodium hydroxide.
3. Determine the pH at the equivalence point if
15.0 mL of 0.20 M nitric acid is titrated with
0.20 M ammonia.
70
Practice Problems
4. a) Calculate the pH of a solution made by
mixing 50.0 mL of 0.15 M formic acid and
0.41 g of sodium formate.
b) Calculate the pH if 10.0 mL of 0.10 M NaOH
is added.
5. What is the pH of a solution made by mixing
25.0 mL of 0.20 M benzoic acid and 45.0 mL of
0.10 M sodium benzoate?
6. How many moles of sodium carbonate must
be added to 0.20 mole sodium hydrogen
carbonate in 250. mL to obtain a pH of 10.00?
71
Practice Problems
7. How many mL’s of 0.30 HCl must be added to
25.0 mL of 0.500 M sodium phosphate to
produce a solution with a pH of 13.00?
8. A 20.00 mL sample of 0.500 M HNO3 is titrated
with 0.500 M KOH. Calculate the pH of the
solution:
a) before the titration begins.
b) when 10.00 mL of base have been added.
c) when 19.00 mL of base have been added.
d) at the equivalence point.
e) when 21.00 mL of base have been added.
72
Practice Problems
9. A 20.00 mL sample of 0.5000 M formic acid
is titrated with 0.500 M sodium hydroxide.
Calculate the pH of the solution:
a) before the titration begins.
b) when 10.00 mL of base have been added.
c) when 19.00 mL of base have been added.
d) at the equivalence point.
e) when 21.00 mL of base have been added.
73
Practice Problems
10. A 20.00 mL sample of 0.400 M ammonia is
titrated with 0.200 M HCl. Calculate the pH
of the solution:
a) before the titration begins.
b) when 20.00 mL of acid have been added.
c) when 39.00 mL of acid have been added.
d) at the equivalence point.
e) when 41.00 mL of acid have been added.
74
Practice Problems Answers
1. 1.30
2. 11.23
3. 5.12
4. 3.66, 3.77
5. 4.15
6. .096
7. 9.1
8. .301, .77, 2.0, 7.00, 12.00
9. 2.03, 3.74, 5.03, 8.57, 12.0
10. 11.43, 9.25, 7.66, 5.07, 2.48
The End!!
Acid-Base Reactions
1. Calculate the pH if 25.0 mL 0.20 M of HCl is
added to 40.0 mL of 0.20 M NaOH.
HCl + NaOH ---> NaCl + HOH
5.0
8.0
0
- 5.0
- 5.0
+ 5.0
3.0
5.0
0
[OH-] =
3.0
= 0.046 M
65.0
pOH = 1.34 pH = 12.66
75
Acid-Base Reactions
2. Calculate the pH at the equivalence point
if 25.00 mL 0.20 M of HCl is titrated with
0.20 M NH3.
HCl + NH3 ---> NH4+ + Cl5.0
5.0
0
- 5.0
- 5.0
+ 5.0
0
[NH4+] =
0
5.0
50.
5.0
= 0.10 M
0
76
Acid-Base Reactions
77
2. [NH4+] = 0.10 M
NH4+ <---> NH3 + H+
O.10
0
0
-x
+x
+x
0.10
x
x
Ka =
[H+][NH3]
[NH4+]
=
X = 7.5 x 10-6 M = [H+]
x2
(0.10)
= 5.6 x 10-10
pH = 5.12
Acid-Base Reactions
3. Calculate the pH at the equivalence point if
30.00 mL 0.20 M of HC2H3O2 is titrated with
0.30 M NaOH.
HC2H3O2 + OH- ---> HOH + C2H3O26.0
6.0
0
- 6.0
- 6.0
+ 6.0
0
[C2H3O2- ]=
0
6.0
50.
6.0
= 0.12M
78
79
Acid-Base Reactions
3. [C2H3O2- ]= 0.12 M
C2H3O2- + HOH <---> HC2H3O2+ OH-
0.12
0
-x
+x
+x
x
x
0.12
Kb =
[HC2H3O2][OH-]
[C2H3O2-]
=
X = 8.2 x 10-6 = [OH-]
0
x2
(0.12)
= 5.6 x 10-10
pH = 8.91
Sample Problem
80
Calculate the pH of a 0.100 M HC2H3O2
solution.
HC2H3O2 <--->
H+ + C2H3O20.100
-x
0
0
+x
+x
x
x
0.100
Ka =
[C2H3O2-][H+]
[HC2H3O2]
=
X = 1.3 x 10-3 M = [H+]
x2
(0.100)
= 1.8 x 10-5
pH = 2.89
Sample Problem
81
Calculate the pH of a solution that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2
HC2H3O2 <---> H+ + C2H3O20.100
-x
0
+x
0.100
Ka =
[C2H3O2-][H+]
[HC2H3O2]
x
=
X = 1.8 x 10-5 M = [H+]
0.100
+x
0.100
x(0.100)
(0.100)
= 1.8 x 10-5
pH = 4.74
Sample Problem
1.00 L H2O has a pH = 7.00
Calculate the pH if 0.010 mole HCL is added.
[H+
]=
.010
= 0.010
1.00
pH = 2.00
Adding 0.010 mole HCl changes the
pH from 7.00 to 2.00, 5.00 pH units.
82
Sample Problem
83
1.00 L 0.100 M HC2H3O2 and 0.100 M NaC2H3O2
has a pH = 4.74
Calculate the pH if 0.010 mole HCL is added.
C2H3O2- + H+ ---> HC2H3O2
0.100
- 0.010
0.090
[C2H3O2- ] =
0.010
0.100
- 0.010
+ 0.010
0
0.110
.090
= 0.090 M
1.00
.110
[HC2H3O2 ] =
= 0.110 M
1.00
Sample Problem
HC2H3O2
<--->
0.110
-x
0.110
Ka =
[C2H3O2-][H+]
[HC2H3O2]
H+
+
C2H3O2-
0
0.090
+x
x
+x
0.090
=
X = 2.2 x 10-5 M = [H+]
84
x(0.090)
(0.110)
= 1.8 x 10-5
pH = 4.66
Adding 0.010 mole HCl changes the pH
from 4.74 to 4.66, only 0.08 pH units.
Preparing a Buffer
85
Preparing Buffers
1. Solid/Solid: mix two solids. (Example 1)
2. Solid/Solution: mix one solid and one
solution.(Example 2)
3. Solution/Solution: mix two solutions.
(Example 3)
4. Neutralization: Mix weak acid with
strong base (Examples 4 and 5)
or weak base with strong acid.
(Example 6)
Sample Problem
86
1. Calculate the pH of a solution made by mixing
1.5 moles of phthalic acid and 1.2 moles of
sodium hydrogen phthalate in 500. mL of sol’n.
Ka= 3.0 x 10-4 Conjugates
HA
<--->
H+
+do not
A- react!!
3.0
0
2.4
-x
+x
+x
3.0 - x
x
2.4 + x
Ka =
[A-] [H+]
[HA]
=
(2.4 + x)(x)
(3.0 + 0)
x = 3.7 x 10-4 M
= 3.0 x 10-4
pH = 3.43
Sample Problem
87
2. How many moles of sodium acetate must be
added to 500. mL of 0.25 M acetic acid to
produce a solution with a pH of 5.50?
X = moles NaC2H3O2Conjugates
HC2H3O2 <---> H+ + C2H3O20.25
0
- 3.2x10-6
0.25
Ka =
do not react!!
x/0.500
+ 3.2x10-6 + 3.2x10-6
3.2x10-6
x/0.500
[C2H3O2-][H+]
[HC2H3O2]
=
(x/0.500)(3.2x10-6 )
(0.25)
x = 0.70 mole NaC2H3O2
= 1.8
x 10-5
Sample Problem
3. How many mLs of 0.10 M sodium acetate
must be added to 20.0 mL of 0.20 M acetic
acid to produce a solution with a pH of 3.50?
X = mLs NaC2H3O2
HC2H3O2
<--->
0.20(20.0/20.0+x)
- 3.2x10-4
0.20(20.0/20.0+x)
H+
C2H3O2-
+
0
0.10(x/20.0+x)
+ 3.2x10-4
3.2x10-4
Conjugates
do not react!!
+ 3.2x10-4
0.10(x/20.0+x)
88
Sample Problem
Ka =
[C2H3O2-][H+]
[HC2H3O2]
{0.10(x/20.0+x)}(3.2x10-4 )
0.20(20.0/20.0+x)
x = 2.2 mL NaC2H3O2
= 1.8 x 10-5
89
Sample Problem
4. How many moles of potassium hydroxide
must be added to 500. mL of .250 M HCN to
produce a solution with a pH of 9.00?
X = moles KOH
OH- + HCN ---> CN- + HOH
x
-x
0
[CN-] =
0.125
0
-x
+x
0.125 - x
x
x
.500
0.125 - x
[HCN] =
0.500
90
Sample Problem
HCN
<--->
H+
+
(0.125 - x)/0.500
- 1.0 x 10-9
[CN-][H+]
[HCN]
CNx/0.500
+ 1.0 x 10-9 + 1.0 x 10-9
(0.125 - x)/0.500
Ka =
91
1.0 x 10-9
x/0.500
x/.500 (1.0 x 10-9)
=
(.125 - x)/.500)
X = 0.036 mole KOH
= 4.0x10-10
Sample Problem
92
5. How many mLs of 0.30 M sodium hydroxide
must be added to 25.0 mL of 0.500 M acetic acid
to produce a solution with a pH of 4.10?
X =mL NaOH
HC2H3O2 + OH- ---> HOH + C2H3O212.5
0.30x
0
- 0.30x - 0.30x
+ 0.30x
12.5 - 0.30x
[C2H3O2- ] =
0
0.30x
25.0 + x
0.30x
[HC2H3O2] =
12.5 - 0.30x
25.0 + x
Sample Problem
HC2H3O2
<--->
(12.5 - 0.30x)/(25.0+x)
- 7.9 x10-5
+
H+
0.30x/(25.0+x)
0
+ 7.9 x10-5
(12.5 - 0.30x)/(25.0+x)
Ka =
C2H3O2-
[C2H3O2-][H+]
[HC2H3O2]
93
+ 7.9x 10-5
0.30x/(25.0+x)
7.9 x 10-5
{0.30x/(25.0+x)} 7.9x 10-5
=
(12.5 - 0.30x)/(25.0+x))
= 1.8 x 10-5
X = 7.6 mL NaOH
Sample Problem
6. Calculate the pH of a solution made by
mixing 50.0 mL of 0.15 M NH3 and 20.0 mL
of 0.10 M HCl.
HCl + NH3 ---> NH4+ + Cl2.0
7.5
0
0
- 2.0
- 2.0
+ 2.0
NA
0
5.5
2.0
NA
[NH4+] =
2.0
= 0.029 M
70.0
[NH3] =
5.5
70.0
= 0.079 M
94
Sample Problem
NH4+
Ka =
<--->
NH3
H+
+
0.029
0.079
0
-x
+x
+x
0.029
0.079
[H+][NH3]
[NH4+]
=
x
x(0.079)
(0.029)
X = 2.1 x 10-10 M = [H+]
= 5.6 x 10-10
pH = 9.68
For more practice do all General problems 84-101
95
Sample Problems
1. Calculate the pH of a 0.10 M
HNO2 solution before and after
making the solution 0.25 M in
NaNO2.
96
Sample Problem
97
Calculate the pH of a 0.10 M HNO2
solution.
HNO2 <---> H+ + NO20.10
-x
0.10 - x
Ka =
[NO2-][H+]
[HNO2]
0
0
+x
+x
x
x
=
X = 6.5 x 10-3 M = [H+]
x2
(.10 - x)
= 4.5 x 10-4
pH = 2.19
Sample Problem
98
Calculate the pH of a 0.10 M HNO2 and 0.25 M
NaNO2 solution.
HNO2 <---> H+ + NO20.10
-x
0.10 - x
Ka =
[NO2-] [H+]
[HNO2]
0
0.25
+x
+x
x
=
X = 1.8 x 10-4 M = [H+]
0.25 + x
x(0.25 + x)
(0.10 - x)
= 4.5 x 10-4
pH = 3.74
Sample Problems
2. Calculate the pH of 0.500 L of
a buffer solution composed of
0.10 M sodium acetate and
0.15 M acetic acid, before and
after adding 1.0 grams of
sodium hydroxide solid
(no volume change).
99
Sample Problem
100
Calculate the pH of 0.500 L of a buffer
solution composed of 0.10 M sodium
acetate and 0.15 M acetic acid.
HC2H3O2 <---> H+ + C2H3O2-
Ka =
0.15
0
-x
+x
0.15
x
[C2H3O2-] [H+]
[HC2H3O2]
=
X = 2.7 x 10-5 M = [H+]
0.10
+x
0.10
x(0.10)
(0.15)
= 1.8 x 10-5
pH = 4.57
Sample Problem
101
Calculate the pH after adding 1.0 grams
of sodium hydroxide solid.
HC2H3O
NaC2H3O
NaOH
(.500L)(.15M) = .075 mole
(.500L)(.10M) = .050 mole
(1.0g)(40.0g/mole) = .025 mole
HC2H3O2 + OH- ---> HOH + C2H3O20.075
0.025
0.050
- 0.025
- 0.025
+ 0.025
0.050
[HC2H3O2] =
.050
.500
0
0 .075
-]
= 0.10 [C2H3O2 =
.075
.500
= 0.15 M
Sample Problem
102
Calculate the pH after adding 1.0 grams
of sodium hydroxide solid.
HC2H3O2
<--->
H+
+
C2H3O2-
0.10
Ka =
0.15
-x
+x
+x
0.15
x
0.10
[C2H3O2-][H+]
[HC2H3O2]
=
X = 1.2 x 10-5 M = [H+]
x(0.15)
(0.10)
= 1.8 x 10-5
pH = 4.92
Sample Problems
3. Calculate the pH of a solution
that is 0.18 M in Na2HPO4 and
0.12 M in NaH2PO4.
103
Sample Problem
H2PO4-
Ka =
<--->
H+
+
HPO42-
0.12
0
0.18
-x
+x
+x
0.12
x
0.18
[HPO42-][H+]
[H2PO4-]
=
X = 4.1 x 10-8 M = [H+]
104
x(0.18)
(0.12)
= 6.2 x 10-8
pH = 7.38
Sample Problems
105
4. Suggest an appropriate buffer
system for pH 5.0.
Name of Acid
Oxalic Acid
Hydrogen Sulfate Ion
Phosphoric Acid
Formic Acid
Hydrogen Oxalate Ion
Acetic Acid
Dihydrogen Phosphate Ion
Boric Acid
Ammonium Ion
Hydrogen Carbonate Ion
Hydrogen Phosphate Ion
Ka
pKa
3.8 x 10-2
1.2 x 10-2
7.1 x 10-3
1.8 x 10-4
5.0 x 10-4
1.8 x 10-5
6.3 x 10-8
6.0 x 10-10
9.6 x 10-10
4.7 x 10-11
4.4 x 10-13
1.42
1.92
2.15
3.74
4.30
4.74
7.20
9.22
9.25
10.33
12.36