Problem 1
What is the pH of a 291mL sample of 2.993M benzoic acid (C6H5COOH)
(Ka=6.4x10-5)?
Write out acid dissociation reaction:
C6H5COOH ↔ C6H5COO- + H+
Make an ICE chart since this is a weak acid equilibrium:
+] - + H[+
C6H5[COOH ↔ C6H[H
5COO
I
C
2.993
-x
0
+x
0
+x
E
2.993 - x
x
x
Write out Ka and solve:
[C6 H 5COO- ][H  ]
Ka 
[C6 H 5COOH]
x2
6.4 10 
2.993
x  0.01384M  [H  ]
-5
pH  log(0.0138 4M )
pH  1.86
Problem 2
A 489mL sample of 0.5542M HNO3 is mixed with 427mL sample of NaOH (which
has a pH of 14.06). What is the pH of the resulting solution?
Write out the reaction:
H+ + NO3- + Na+ + OH- → Na+ + NO3- + H2O
Or after eliminating spectator ions:
H+ + OH- → H2O
Now, make an ICE chart using moles to find out what’s left after reaction:
What’s [NaOH]?
H[ +
+ OH[ - → H2[O
I 0.271
C - 0.271
0.490
- 0.271
0
+ 0.271
E
0.2193
0.271
0
Now solve for pOH using moles of
0.2193moles
[OH- ] 
(0.489  0.427)L
[OH ]  0.2394M
pOH  0.621
pH  13.38
OH-,
and then find pH:
pOH  14  14.06
pOH  0.06
[OH- ]  100.06
[OH- ]  1.148
Problem 3
The Ka values at 25oC for a series of acids is given:
1.8 x 10-5
7.6 x 10-4
1.3 x 10-3
1.4 x 10-3
2.2 x 10-2
Which of the following acids has a Ka = 1.4 x 10-3?
First realize all the answers are forms of
ethanoic acid.
Then, realize they want you to pick the 2nd
strongest acid of the bunch since it has
the 2nd largest Ka.
So you want to analyze each acid’s
conjugate base stability:
Notice that the electronegative halogens
act to stabilize the conjugate base, which
come from the strongest acids. Cl is most
electronegative, and most stabilizing.
c
Problem 4
Consider a solution containing 0.45M HCN and 0.69M NaCN for the next two
questions. The Ka for HCN = 6.2 x 10-10.
What is the pH of this solution?
Write out the reaction of the acid dissociation:
HCN ↔ H+ + CNWrite out the ions that the salt produces:
NaCN → Na+ + CNNow, make a chart using molarity to find out what’s in this buffer solution:
I
HCN
[
0.45
↔
H[ +
0
+
CN
[
0.69
Now solve for pH using Henderson-Hasselbalch
equation:

 [CN ] 

pH  pKa  log
 [HCN] 
 0.69M 
 log(6.2 10 10 )  log

0.45M


 9.208  0.186
pH  9.39
Problem 5
Consider a solution containing 0.45M HCN and 0.69M NaCN for the next two
questions. The Ka for HCN = 6.2 x 10-10.
Now calculate the pH of this solution after 0.25 mol of NaOH is added to 1.00L of
the solution. Assume that the volume does not change.
The added NaOH (base) will react with the HCN (acid), so we should form an ICE
chart to see what remains after reaction:
I
C
HCN
+ OH- ↔ [H+H]2O
[
0.45
0.25
-- 0.25
- 0.25
--
E
0.20
0
--
+ CN
[ 0.69
+ 0.25
0.94
Now solve for pH using Henderson-Hasselbalch equation:
 [CN  ] 

pH  pKa  log
 [HCN] 
 0.94moles 
 log(6.2 10 10 )  log

0.20moles


pH  9.88
Problem 6
Consider a 7.3 x 10-10M solution of HCl at 285 K. What is the pH of the solution?
Since this is a strong acid, we can simply take [HCl]=[H+] since it fully dissociates,
and find pH easily:
pH  -log (7.3 10-10 M)
pH  9.14
Does this make any sense? What’s wrong with it?
Think about it. We’re calculating pH of an acid…yet pH>7. What does this mean?
Well, look at the concentration of HCl; it is extremely tiny, so it is negligible. In
other words, when compared to water, HCl would lose in this case since it’s so
dilute.
Simply pH is simply that of neutral water (pH=7)!
Problem 7
For the following reaction, K < 1 at room temperature.
HOCl (aq) + HCO3- (aq) ↔ H2CO3 (aq) + OCl- (aq)
Which of the following statements is true?
A) H2CO3 is a stronger acid than HOCl. True. Since K<1, we know that the
stronger acid is on the product side (H2CO3).
B) HOCl is a stronger acid than H2CO3. False. See A.
C) OCl- is a stronger acid than HOCl. False. OCl- is a base.
D) HCO3- is stronger base than OCl-. False. Since K<1, we know the stronger
base is on the product side (OCl-).
E) More information is needed.
Problem 8
What is the pH of a solution that results from adding 0.081 mol Ca(OH)2 (s) to
1.37L of a buffer comprised of 0.21M HF and 0.21M NaF?
First, write out the buffer solution’s reaction:
HF ↔ H+ + FNow, if we add Ca(OH)2 (strong base), it will react with the HF (acid). So let’s
construct an ICE table in moles to find out what’s left after reaction:
I
C
HF
[ + OH- ↔[H+H]2O + F[ 0.2877 2(0.081)
-- 0.2877
- 0.162
- 0.162
-- + 0.162
E
0.1257
0
--
0.4497
Now solve for pH using Henderson-Hasselbalch equation:
 [F ] 

pH  pKa  log
 [HF] 
 0.4497moles 
 log(7.2 10 4 )  log

0.1257mole
s


pH  3.70
Notice how Ca(OH)2
will dissociate to form
twice as many (OH-).
Problem 9
A 0.251L solution of 1.89M sulfurous acid (Ka1 = 1.5x10-2 and Ka2 = 1.0x10-7) is
titrated with 1.25M NaOH.
What will the pH of the solution be when 0.6793L of the NaOH has been added?
First, write out the first acid dissociation: (notice how you don’t need to know
what sulfurous acid is, just keep in mind it’s diprotic)
H2A ↔ HA- + H+
Now, if we add NaOH (strong base), it will react with the H2A (acid). So let’s
construct an ICE table in moles to find out what’s left after reaction:
H2[A
I 0.4744
C - 0.4744
E
+
↔[H+H]2O + HA
[
0.8491
-0
- 0.4744
-- + 0.4744
0
OH-
0.3747
--
0.4744
Are we done? No. Remember, this is diprotic and so we have excess OH- that will
react with HA-.
…continued on next slide….
Problem 9
A 0.251L solution of 1.89M sulfurous acid (Ka1 = 1.5x10-2 and Ka2 = 1.0x10-7) is
titrated with 1.25M NaOH.
What will the pH of the solution be when 0.6793L of the NaOH has been added?
… continued from previous slide…
Now, write out the second acid dissociation:
HA- ↔ A2- + H+
Now, if we have extra OH- (strong base), it will react with the HA- (acid). So let’s
construct an ICE table in moles to find out what’s left after reaction:
HA
+ OH- ↔[H+H] 2O + [A2[
I 0.4744
0.3747
-0
C - 0.3747 - 0.3747
-- + 0.3747
E 0.0997
0
-- 0.3747
Now use Henderson-Hasselbalch to get the pH:
 [A 2 ] 

pH  pKa2  log
- 
 [HA ] 
 0.3747moles 
 log(1.0 10 7 )  log

 0.0997moles 
pH  7.58
Problem 10
The titration of 1.00L of a 1.00M solution of the triprotic acid, H3A, with 1.00M
NaOH is shown below. It is not drawn to scale.
What is the pH at D?
At point D, we are at the 2nd
halfway point. So we should use
the following formula:
pH  pKa2
 log(6.43 10 7 )
pH  6.19
H3A + H2O ↔ H3O+ + H2AH2A- + H2O ↔ H3O+ + HA2HA2- + H2O ↔ H3O+ + A3-
Ka1=3.29x10-3
Ka2=6.43x10-7
Ka3=4.25x10-10
Problem 11
The titration of 1.00L of a 1.00M solution of the triprotic acid, H3A, with 1.00M
NaOH is shown below. It is not drawn to scale.
What is the pH at C?
At point C, we are at the 1st
equivalence point. So we should
use the following formula:
pKa1  pKa2
2
 log(3.29 10 3 )  log(6.43 10 7 )

2
pH  4.33
pH 
H3A + H2O ↔ H3O+ + H2AH2A- + H2O ↔ H3O+ + HA2HA2- + H2O ↔ H3O+ + A3-
Ka1=3.29x10-3
Ka2=6.43x10-7
Ka3=4.25x10-10
Problem 12
Consider a 0.35M solution of each of the following salts in distilled water. Will
the solution be acidic, neutral, or basic?
You may need to reference a table of Ka and Kb Tables values. .
NH4OBr:
 NH4+ + H2O → NH3 + H3O+
Ka=5.6x10-10
Kb > Ka  Basic
 OBr- + H2O → HOBr + OHKb=5.0x10-6
BaClO4:
 Ba2+ + H2O → no reaction (from strong base)
 ClO4- + H2O → no reaction (from strong acid)
 Neutral
RbCH3COO:
 Rb+ + H2O → no reaction (from strong base)
 CH3COO- + H2O → CH3COOH + OH-
 Basic
(CH3CH2)3NHNO2:
 (CH3CH2)3NH+ + H2O → (CH3CH2)3N + H3O+
 NO2- + H2O → HNO2 + OH-
Ka=2.5x10-11
Kb=2.5x10-11
Kb = Ka
 Neutral
Problem 13
How many of the following are true?
1. The buffer with the greatest buffering capacity for a phenol/sodium
phenolate buffer system will have a pH of 9.60. False, phenol: Ka=1.6x10-10, so
pH=pKa=9.80.
2. A buffer resists pH change when base is added by converting hydroxide ions
to hydronium ions.. False, the hydroxide (OH-) would react with the acid (HA)
to form H2O, not H3O+.
3. A buffered system can be formed by combining a weak acid/conjugate base
pair in solution. True, this is a weak acid+salt combination.
4. A buffer is formed when you add HCl to NH3. True, but only near the halfway
point.
5. A buffer is formed when you add HCl to CH3COOH. False, this is a weak acid
and strong acid addition, pH would just increase a lot.
Problem 14
Which of the following solutions will produce a buffer with pH near 10.50?
1.
1.00L of 0.50M HSO4- (Ka = 1.0x10-7) + 1.00L of 0.25M KOH.
No, pH=pKa=7.
2.
2.00L 0.50M H2NNH2 (Kb = 3.0x10-6) + 2.00L 0.25M HI.
No, Ka=3.33x10-9, so pH=pKa=8.50.
3.
0.100L 1.00M H3BO3 (Ka = 5.8x10-10) + 0.020L 2.50M HBr.
No, pH=pKa=9.20.
4.
0.100L of 0.50M HONH2 (Kb = 1.1x10-8) + 0.50L of 0.050M HNO3.
No, Ka=9.1x10-7, so pH=pKa=6.04.
5.
0.500L 1.00M CH3NH2 (Kb = 4.4x10-4) + 1.00L 0.25M HI.
Yes, Ka=2.27x10-11, so pH=pKa=10.64.
Problem 15
The next three problems deal with the titration of 145mL of 1.35M methylamine
CH3NH2 (Kb = 4.4 x 10-4) with 0.25M HCl.
Water will be a major species throughout the titration. The chemical
species, in addition to water, that can be found in this reaction mixture
during the titration are:
I. H+ II. OH- III. Cl- IV. CH3NH2 V. CH3NH3+
What is the pH at equivalence point?
At equivalence point, we want equal moles of CH3NH2 and HCl:
I
CH[3NH2 + H+ ↔ CH3NH3+
0.196
0.196
0
C
E
- 0.196
0
- 0.196
0
+ 0.196
0.196
Now, we want to calculate the pH of CH3NH3+ in water:
Remember to use molarities, so divide by total volume
0.145L from methylamine
0.784L from HCl (you get this from 0.196moles/0.25M=0.784L)
0.929L Total Volume
…continued on next slide…
Problem 15
The next three problems deal with the titration of 145mL of 1.35M methylamine
CH3NH2 (Kb = 4.4 x 10-4) with 0.25M HCl.
Water will be a major species throughout the titration. The chemical
species, in addition to water, that can be found in this reaction mixture
during the titration are:
I. H+ II. OH- III. Cl- IV. CH3NH2 V. CH3NH3+
What is the pH at equivalence point?
…continued from previous slide…
So, we should write out the ion’s (it’s an acid) reaction in water: [use molarities!]
CH3NH
[ 3+ +
H2O ↔ CH3NH2
+
H3O+
I
0.211
--
0
0
C
E
-x
0.211 – x
---
+x
x
+x
x

First, find Ka:
1.0 10-14
Ka 
4.4 10-4
K a  2.27 10 11
[CH 3 NH3 ][OH ]
Ka 
[CH 3 NH2 ]
x2
2.27 10 
0.211
x  2.19 10 6 M  [H 3O  ]
-11
pH  log(2.19 10-6 M)
pH  5.66
Problem 16
The next three problems deal with the titration of 145mL of 1.35M methylamine
CH3NH2 (Kb = 4.4 x 10-4) with 0.25M HCl.
Water will be a major species throughout the titration. The chemical
species, in addition to water, that can be found in this reaction mixture
during the titration are:
I. H+ II. OH- III. Cl- IV. CH3NH2 V. CH3NH3+
How many mL of HCl will need to be added to reach pH=10.64?
First, realize from 18 that you are before equivalence point, so use HendersonHasselbalch equation:
This is the halfway point!

 [CH 3 NH3 ] 
So we know the number of

pOH  pKb  log

 [CH 3 NH 2 ] 
moles of HCl is half that of
 [CH 3 NH3  ] 
base CH3NH2! n  1 n
4


3.36  log(4.4 10 )  log
HCl
CH3 NH 2

2
 1.35M 
1
 [CH 3 NH3  ] 
  0.196moles

 3.36  log

2
 1.35M 
n HCl  0.098moles
 [CH 3 NH3  ] 

0  log

0.098 moles
 1.35M 
VHCl 

0.25 M
[CH 3 NH3 ]
1
VHCl  0.392L
1.35M
Problem 17
The next three problems deal with the titration of 145mL of 1.35M methylamine
CH3NH2 (Kb = 4.4 x 10-4) with 0.25M HCl.
Water will be a major species throughout the titration. The chemical
species, in addition to water, that can be found in this reaction mixture
during the titration are:
I. H+ II. OH- III. Cl- IV. CH3NH2 V. CH3NH3+
What are the major species when 783mL of HCl has been added?
Let’s write out the ICE chart in moles to see what’s left after reaction:
I
CH[3NH2 + H+ ↔ CH3NH3+
0.196
0.196
0
C
E
- 0.196
0
- 0.196
0
+ 0.196
0.196
So this is equivalence point!
We only have CH3NH3+ and Cl- (unreacted) in solution
Problem 18
A 0.613M solution of a weak acid has a pH of 2.670 at 309K. What is the Ka of
the weak acid?
Write out acid dissociation reaction (generic is fine):
HA ↔ A- + H+
Make an ICE chart since this is a weak acid equilibrium:
HA
[
↔
A+- ] +
[H
H[+
First, find x from pH:
 10  2.670
I
C
0.613
-x
0
-x
0
-x
E
0.613 - x
x
x
Write out Ka (plug in x) and solve:
[A - ][H  ]
Ka 
[HA]
x2

0.613
(0.00214) 2

0.613
K a  7.4 10 6
x  [H  ]
x  0.00214M
Problem 19
Consider the following titration curve:
How many of the following are
true?
1. The pH at point B is pKa1.
True, this is the 1st halfway.
2. The pH at point E will be the
average of pKa1 and pKa2.
False, you want pKa2 and pKa3
since it’s the 2nd eq. pt.
3. To the right of point F
[A3-] > [HA2-].
True, it’s after the 3rd
halfway point.
4. HA2- will be a major species
at point E.
True, you have lost all H2A-.
5. The second dissociation of H+
Which is based on the following reactions:
will play a large role in the
H3A ↔ H+ + H2ApH level at point E.
H2A- ↔ H+ + HA2False, only the first
HA2- ↔ H+ + A3dissociation matters.