Problem 1 Exam I
What is the pH of a 0.443 M solution of LiOH? Since LiOH Is a strong base, we know that it dissociates completely in water:
LiOH Li+ + OH–
So we can calculate pOH simply: And then pH from the pOH
pOH= -log[OH-]
= -log(0.443)
pOH= 0.35
pH=14-pOH
=14-0.35
=13.65
Problem 2 Exam I
What is the pH of a 289 mL sample of 2.453 M acetic acid (CH3COOH)? (Ka = 1.8 × 10-5)
Write out acid dissociation reaction: CH3COOH ↔ CH3COO– + H+
Make an ICE chart since this is a weak acid equilibrium: Write out Ka and solve:
I 2.453 0 0
C - x + x + x
E 2.453 - x x x
CH3COOH ↔ CH3COO– + H+
Ka=[CH
3COO- ][H+ ]
[CH3COOH]
1.8´10-5 »x2
2.453
x = 0.00664M= [H+ ]
pH= -log(0.00664)
pH= 2.18
NOTE: Normally, I’d tell you to do your ICE chart in moles, but since there’s no volume change here, it doesn’t matter. You’d get the same answer whether you do this in moles or M.
Problem 3 Exam I
To a 1.00 L buffer solution made of 2.01 M formic acid (HCOOH) and 1.23 M potassium formate (KOOCH) was added 0.549 moles of NaOH (assume no volume change to the solution). What is the final pH of this buffer assuming Ka (formic acid) = 1.80 x 10–4? Write out the reaction of the acid dissociation: HCOOH ↔ H+ + HCOO– weak acid Write out the ions that the salt produces: + KOOCH → K+ + HCOO– salt (conj. base)
The fact that we have a weak acid and its conjugate base (which comes from the salt) means that we have a buffer solution. Now, make a chart using moles to find out what’s in this buffer: Now we can introduce the the 0.549 moles of NaOH, which will react with the acid:
…continued on next slide…
I 2.01 0 1.23
HCOOH ↔ H+ + HCOO–
Problem 3 Exam I
To a 1.00 L buffer solution made of 2.01 M formic acid (HCOOH) and 1.23 M potassium formate (KOOCH) was added 0.549 moles of NaOH (assume no volume change to the solution). What is the final pH of this buffer assuming Ka (formic acid) = 1.80 x 10–4? Now we can introduce the the 0.549 moles of NaOH, which will react with the acid: ICE chart in moles Now solve for pH using the Henderson-Hasselbach equation:
…continued from previous slide…
I 2.01 0.549 1.23 —
C -0.549 -0.549 +0.549 —
E 1.461 0 1.779 —
HCOOH + OH– ↔ HCOO– + H2O
a
4
[HCOO ]pH pK log
[HCOOH]
1.779 mollog(1.80 10 ) log
1.461 mol
pH 3.830
NOTE: You should really plug in molarities into the H-H equation, but since the volumes are the same, it doesn’t matter as they cancel out anyway.
Problem 4 Exam I
Calculate the pH of a 1.15 M solution of NaNO2. First, you should notice that NaNO2 is a salt and it dissociates into its ions:
NaNO2 (s) Na+ (aq) + NO2– (aq)
Each of these ions reacts with H2O: Na+ (aq) + H2O (l) no reaction NO2
– (aq) + H2O (l) HNO2 (aq) + OH– (aq) Since only NO2
– reacts with H2O, this is a basic salt. And we can set up the weak base equilibrium: Write out Kb and solve for x: Use that x to solve for pOH and pH:
I 1.15 — 0 0
C - x — + x + x
E 1.15 - x — x x
NO2– + H2O ↔ HNO2 + OH–
b
2-1
2
6
1
2[HNO ][OH ]K
[NO ]
x2.50 10
1.15
x 5.362 10 M [OH ]
6pOH log(5.362 10 )
pOH 5.271
pH 8.729
Problem 5 Exam I
Consider the titration of 1.0 M carbonic acid (H2CO3) with 2.0 M NaOH. What is the pH at the second halfway point of the titration? First, you should realize that H2CO3 is a polyprotic (diprotic) acid, meaning it dissociates twice: H2CO3 ↔ HCO3
– + H+ Ka1 = 4.3 × 10–7
HCO3– ↔ CO3
2– + H+ Ka2 = 5.6 × 10–11
The halfway point is defined at the point at which the nacid = nconjugate base. Since it’s asking about the second halfway point this is the point at which And at this point, pH = pKa2:
2-3 3HCO CO
n n
a2
11
pH logK
log(5.6 10 )
10.25
Problem 6 Exam I
A 3.914 M solution of diethylamine ((C2H5)2NH, Kb = 1.30×10–3) has a pH of 12.85. What is the percent ionization of this solution? Since this is a weak base, we can write out the weak base equilibrium:
(C2H5)2NH + H2O ↔ (C2H5)2NH2+ + OH–
And the percent ionization will be defined as the concentration of (C2H5)2NH2+
formed relative to the initial amount of (C2H5)2NH. Let’s fill out an ICE chart: They tell us that pH is 12.85, which means we can back-calculate the concentration of OH– from this and that will give us the x. Now, let’s plug it into the ICE chart. …continued on next slide…
I 3.914 — 0 0
C - x — + x + x
E 1.15 - x — x x
(C2H5)2NH + H2O ↔ (C2H5)2NH2+ + OH–
1.15
pOH 14 pH
14 12.85
pOH 1.
] 10
15
[OH
0.708 M
Problem 6 Exam I
A 3.914 M solution of diethylamine ((C2H5)2NH, Kb = 1.30×10–3) has a pH of 12.85. What is the percent ionization of this solution? Since this is a weak base, we can write out the weak base equilibrium:
(C2H5)2NH + H2O ↔ (C2H5)2NH2+ + OH–
And the percent ionization will be defined as the concentration of (C2H5)2NH2+
formed relative to the initial amount of (C2H5)2NH. Let’s fill out an ICE chart (in molarity here since constant volume): Now we can calculate percent ionization:
…continued from previous slide…
I 3.914 — 0 0
C - 0.0708 — + 0.0708 + 0.0708
E 3.843 — 0.0708 0.0708
(C2H5)2NH + H2O ↔ (C2H5)2NH2+ + OH–
2 5 2 2
2 5 2
[(C H ) NH ]% ionization 100 %
[(C H ) NH]
0.0708100 %
3.914
1.81 %
Problem 7 Exam I
Arrange the following 0.6 M solutions in order of increasing pH: NaOH KClO4 HNO3 NaCH3COO HNO2
Firstly, we know that NaOH will have the highest pH since it is the strongest base. Likewise, HNO3 will have the lowest pH since it is the strongest acid. HNO2 is a weak acid (pH < 7). The other two (KClO4 and NaCH3COO) are salts. So we need to determine the relative pH’s of these salts. And now we can arrange this in order as:
HNO3 < HNO2 < KClO4 < NaCH3COO < NaOH
First, dissociate the salt into its ions: KClO4 (s) K+ (aq) + ClO4
– (aq) Each of these ions reacts with H2O: K+ (aq) + H2O (l) no reaction ClO4
– (aq) + H2O (l) no reaction This is a neutral salt: pH = 7
First, dissociate the salt into its ions: NaCH3COO (s) Na+ (aq) + CH3COO– (aq)
Each of these ions reacts with H2O: Na+ (aq) + H2O (l) no reaction CH3COO– (aq) + H2O (l) CH3COOH + OH–
This is a basic salt: pH > 7
Problem 8 Exam I
You find a beaker in your lab labeled “unknown acid”. In an attempt to determine the nature of the acid you determine Ka, which is less than 1.00. Which of the following statements about your acid will be true? 1. The conjugate base of this acid will be a relatively strong weak base. True,
since Ka < 1.00, we know it’s a weak acid so its conjugate base will also be relatively weak.
2. As you dilute the acid, the pH decreases (becomes closer to zero). False, it should increase (go farther from zero).
3. The concentration of OH– will be greater than the concentration of H+ in solution. False, it’s an acid, so [H+] > [OH–].
4. The acid is a strong acid. False, since Ka < 1.00. 5. If titrated with a strong base (like NaOH), the “unknown acid” will have a pH
of 7.0 at the equivalence point. False, this will only be true if it were a strong acid.
Problem 9 Exam I
Calculate how many moles of NaF would be needed to create 1.00 L of a pH = 8.755 solution. Assume the volume does not change as the solution is formed. Notice that we are making a salt solution. So we should split the salt into its ions:
NaF (s) Na+ (aq) + F– (aq) Each of these ions reacts with H2O: Na+ (aq) + H2O (l) no reaction F– (aq) + H2O (l) HF (aq) + OH– (aq) Since only F– reacts with H2O, this is a basic salt. And we can set up the weak base equilibrium ICE chart (in molarity since constant volume): They tell us that pH is 8.755, which means we can back-calculate the concentration of OH– from this and that will give us the x. Now, let’s plug it into the ICE chart. …continued on next slide…
I ??? — 0 0
C - x — + x + x
E ??? - x — x x
F– + H2O ↔ HF + OH–
6]
pOH 5
5
.2
.6
45
[O 88H 0 M1
NOTE: ICE chart is in moles, but since it’s 1.00 L, everything works out anyway if it’s in M.
Problem 9 Exam I
Calculate how many moles of NaF would be needed to create 1.00 L of a pH = 8.755 solution. Assume the volume does not change as the solution is formed. We can set up the weak base equilibrium: So we have x. What we want is the ???, which represents the initial amount of NaF. So let’s set up Kb and solve. But notice that x is small (which is expected), so let’s approximate “??? – x” as just “???”:
…continued from previous slide…
I ??? — 0 0
C - x — + x + x
E ??? - x — x x
F– + H2O ↔ HF + OH– 65.688x M10
b
2-11
6 2
[HF][OH ]K
[F ]
x1.389 10
???
(5.688 10 )
???
??? 2.330 moles =[NaF]
Problem 10 Exam I
How many liters of 0.595 M LiOH will be needed to raise the pH of 0.483 L of 5.69 M sulfurous acid (H2SO3) to a pH of 6.092? First, write out the acid dissociation steps: (keep in mind it’s diprotic)
H2SO3 ↔ HSO3– + H+ Ka1 = 1.5 x 10–2 pKa1 = 1.824
HSO3– ↔ SO3
2– + H+ Ka2 = 1.0 x 10–7 pKa2 = 7.000 This means that in order to get to a pH of 6.092, we need to get past the first equivalence point. We know that the first equivalence point is when the number of moles of acid (H2SO3) and base (OH–) are equal, so:
nH2SO3 = nOH– = 2.748 mol VOH– = 4.618 L Past the first equivalence point, we no longer have any H2SO3; we now have HSO3
– acting as the acid. Now, if we add LiOH (strong base), it will react with the HSO3
– (acid). So let’s construct an ICE table in moles to find out what’s left after reaction: …continued on next slide….
I 2.748 ??? — 0
C - x - x — + x
E 2.748 - x ??? - x — x
HSO3– + OH– ↔ H2O + SO3
–
Problem 10 Exam I
How many liters of 0.595 M LiOH will be needed to raise the pH of 0.483 L of 5.69 M sulfurous acid (H2SO3) to a pH of 6.092? We actually know that we don’t want to hit the second equivalence point; otherwise, the pH will be too high (pKa2 = 7). Using this information, we know that any amount of OH– we add will be limiting because if we add too much it will be in excess and raise the pH > 7. In other words, “??? – x = 0” or “??? = x”. …continued on next slide….
I 2.748 ??? — 0
C - x - x — + x
E 2.748 - x ??? - x — x
HSO3– + OH– ↔ H2O + SO3
–
I 2.748 ??? — 0
C - x - x — + x
E 2.748 - x 0 — x
HSO3– + OH– ↔ H2O + SO3
–
Problem 10 Exam I
How many liters of 0.595 M LiOH will be needed to raise the pH of 0.483 L of 5.69 M sulfurous acid (H2SO3) to a pH of 6.092? Notice now that we have a buffer solution (acid + conjugate base), so we can use the Henderson-Hasselbach equation and solve for x: Remember to add this to the volume we used to get to the first equivalence: VOH– = 4.618 L + 0.508 L = 5.127 L …continued from previous
slide….
I 2.748 ??? — 0
C - x - x — + x
E 2.748 - x 0 — x
HSO3– + OH– ↔ H2O + SO3
–
pH=pKa2
+ log[SO
3
2- ]
[HSO3
- ]
æ
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ö
ø÷
6.092= -log(1.00´10-7)+ logx
2.748- x
æ
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ø÷
x= 0.3023mol=nOH-
VOH-
= 0.508L
Problem 11 Exam I
A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL of water and titrated with 0.41 M NaOH. After 33 mL of the NaOH solution has been added, the overall pH = 5.281. Calculate the Ka value for the HA. Let’s write out the acid dissociation:
HA ↔ A– + H+ Ka=??? Now, if we add NaOH (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: Notice this is a buffer solution (acid + conjugate base). So, now we can use the Henderson-Hasselbalch to get the pKa and Ka:
I 0.035 0.0135 — 0
C - 0.0135 - 0.0135 — +0.0135
E 0.0215 0 — 0.0135
HA + OH– ↔ H2O + A–
pH=pKa+ log
[A- ]
[HA]
æ
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ö
ø÷
5.281= -log(Ka)+ log
0.0135mol
0.0215mol
æ
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ø÷
Ka= 3.3´10-6
Problem 12 Exam I
Consider 1.0 L of a buffered solution based upon HF/NaF where the concentration of both the weak acid and conjugate base are 1.0 M. Choose the statement that is true. 1. A buffer based on HF/NaF would be a good choice to buffer a solution with
a pH of 2.00. False, this buffer would be best for pH = pKa ± 1 = 3.14 ± 1. 2. When 0.5 equivalents of NaOH are added to the buffer the pH will be 3.14.
False, pH = 3.14 would be when no NaOH is added. 3. Adding more NaF to the buffer will decrease the pH. False, this will increase
the pH since this would be adding more conjugate base. 4. Adding 1.0 L of water to the buffer solution will not change the pH of the
solution. True, adding water to the buffer solution will dilute both acid and conjugate base equally so nothing will change. You can also see this in the Henderson-Hasselbach equation since you would dilute both the numerator and denominator by the same factor—it cancels.
5. The system will still resist pH change after 2.0 moles of HCl have been added to the solution. False, we only have 1.0 mol of F– (conjugate base), so adding 2.0 mol of H+ (from HCl) will destroy the buffer.
Problem 13 Exam I
Consider the following titration curve: How many of the following are true? 1. The pH at point G will be the
average of pKa1 and pKa2. True, this is the 1st eq. pt.
2. To the right of point F [A3-] > [HA2-]. True, it’s after the 3rd halfway point.
3. HA2- will be a major species at point C. True, this is the 1st halfway point so we have H3A = H2A-.
4. The pH at G will be determined by the reaction of A3- with water. True, this is the endpoint (we only have A3- and water in solution).
5. The pH at point F is pKa3. True, this is the 3rd halfway point.
Which is based on the following reactions: H3A ↔ H+ + H2A-
H2A- ↔ H+ + HA2-
HA2- ↔ H+ + A3-
Problem 14 Exam I
Consider the following species and rank them from the most acidic (1) to least acidic (7). Which species is ranked 5?
KCl CH3NH2 HNO3 C6H5OH C6H5NH3Cl KOH HCOONa
Firstly, we know that KOH will be the least acidic (7) since it is the strongest base. Likewise, HNO3 will be the most acidic (1) since it is the strongest acid. CH3NH2 is a weak base (pH > 7) and C6H5OH a weak acid (pH < 7). The others (KCl, C6H5NH3Cl, HCOONa) are salts. So we need to determine the relative pH’s of these salts.
First, dissociate the salt into its ions: KCl (s) K+ (aq) + Cl– (aq) Each of these ions reacts with H2O: K+ + H2O no reaction Cl– + H2O no reaction This is a neutral salt: pH = 7
First, dissociate the salt into its ions: C6H5NH3Cl (s) C6H5NH3
+ (aq) + Cl– (aq) Each of these ions reacts with H2O: C6H5NH3
+ + H2O C6H5NH2 + H3O+
Cl– + H2O no reaction
This is an acidic salt: pH < 7
First, dissociate the salt into its ions: HCOONa (s) Na+ (aq) + HCOO– (aq)
Each of these ions reacts with H2O: Na+ + H2O no reaction HCOO– + H2O HCOOH + OH–
This is a basic salt: pH > 7
The species ranked 5 would be one of the basic species: CH3NH2 or HCOONa. We want the weaker base of the two. This would be CH3NH2 based on Kb values.
Problem 15 Exam I
Consider the titration of 0.967 L of 0.893 M ascorbic acid (H2C6H6O6) with 1.72 M NaOH. What is the pH at the second equivalence point of the titration? First, you should realize that H2C6H6O6 is a polyprotic (diprotic) acid, meaning it dissociates twice: H2C6H6O6 ↔ HC6H6O6
– + H+ Ka1 = 7.9 × 10–5
HC6H6O6– ↔ C6H6O6
2– + H+ Ka2 = 1.6 × 10–12
The equivalence point is defined at the point at which the nacid = nbase. Since it’s asking about the second equivalence point, we know we had to go through the first equivalence point which required: Now for the second equivalence point, we also requires 0.8635 mol of OH–—note how we only have HC6H6O6
–. Let’s put this in ICE chart form (in moles):
…continued on next slide…
nH2C6H6O6
=nOH-
= 0.8635mol
I 0.8635 0.8635 0 —
C - 0.8635 - 0.8635 + 0.8635 —
E 0 0 0.8636 —
HC6H6O62– + OH– ↔ C6H6O6
– + H2O
VOH-
= 0.502L
Problem 15 Exam I
Consider the titration of 0.967 L of 0.893 M ascorbic acid (H2C6H6O6) with 1.72 M NaOH. What is the pH at the second equivalence point of the titration? At this point, we have only C6H6O6
2– (weak base) left. And this would be in a weak base equilibrium. Let’s put this in ICE chart form (in molarity, so we should divide by the total volume: 0.967 L + 0.502 L (1st eq. pt.) + 0.502 L (2nd eq. pt) = 1.97 L Write out Kb and solve for x ([OH–], pOH, and pH:
…continued from previous slide…
I 0.4381 — 0 0
C - x — + x + x
E 0.4381 - x — x x
C6H6O62– + H2O ↔ HC6H6O6
– + OH–
Kb
=[HC
6H
6O
6
- ][OH- ]
[C6H
6O
6
2- ]
0.00625»x2
0.4381
x= 0.0523M= [OH- ]
pOH= -log(0.0523)
=1.28
pH=12.72
Problem 16 Exam I
Consider the titration of 135 mL of 1.75 M ethylamine, CH3CH2NH2 (Kb = 5.6×10-4) with 0.25 M HCl. How many mL of HCl will need to be added to ethylamine, CH3CH2NH2, solution to reach a pH of 10.75? First, you should realize that we are very close the halfway point. Let’s write out the ICE chart (in moles): Because we have not gone over the equivalence, any amount of H+ will be limiting. This means that “??? – x =0” or “??? = x”
…continued on next slide…
I 0.236 ??? 0
C - x - x + x
E 0.236 - x ??? - x x
CH3CH2NH2 + H+ ↔ CH3CH2NH3+
I 0.236 ??? 0
C - x - x + x
E 0.236 - x 0 x
CH3CH2NH2 + H+ ↔ CH3CH2NH3+
Problem 16 Exam I
Consider the titration of 135 mL of 1.75 M ethylamine, CH3CH2NH2 (Kb = 5.6×10-4) with 0.25 M HCl. How many mL of HCl will need to be added to ethylamine, CH3CH2NH2, solution to reach a pH of 10.75? Because we have not gone over the equivalence, any amount of H+ will be limiting. This means that “??? – x =0” or “??? = x”
I 0.236 ??? 0
C - x - x + x
E 0.236 - x 0 x
CH3CH2NH2 + H+ ↔ CH3CH2NH3+
Notice now that we have a buffer solution (acid + conjugate base), so we can use the Henderson-Hasselbach equation and solve for x (which represents the amount of H+ we needed to add):
…continued from previous slide….
pOH=pKb+ log
[CH3CH
2NH
3
+ ]
[CH3CH
2NH
2]
æ
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ö
ø÷
3.25= -log(5.6´10-4)+ logx
0.236mol- x
æ
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ø÷
x= 0.118mol=nH+
VH+
= 0.472L
Problem 17 Exam I
A buffer solution based upon sulfurous acid (H2SO3) was created by treating 1.00 L of a 1.00 M sulfurous acid solution with NaOH until a pH of 1.666 was achieved. To this buffer 1.420 moles of NaOH was added. What is the final pH of this solution? (Assume no volume changes) First, write out the first acid dissociation: (keep in mind it’s diprotic)
H2SO3 ↔ HSO3– + H+ Ka1 = 1.5 x 10–2 pKa1 = 1.824
Now let’s figure out the composition of the buffer solution using the Henderson-Hasselbach equation knowing that this reaction produces a buffer. We can set up an ICE chart (in molarity since constant volume) to see what’s left after reaction: …continued on next slide….
pH=pKa1
+ log[HSO
3
- ]
[H2SO
3]
æ
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ö
ø÷
1.666 = -log(1.5´10-2)+ logx
1.00- x
æ
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ø÷
x= 0.410M
I 1.00 ??? — 0
C - x - x — + x
E 1.00 - x 0 — x
H2SO3 + OH– ↔ H2O + HSO3–
Problem 17 Exam I
A buffer solution based upon sulfurous acid (H2SO3) was created by treating 1.00 L of a 1.00 M sulfurous acid solution with NaOH until a pH of 1.666 was achieved. To this buffer 1.420 moles of NaOH was added. What is the final pH of this solution? (Assume no volume changes) Now, if we add NaOH (strong base), it will react with the H2SO3 (acid). So let’s construct an ICE table in moles to find out what’s left after reaction: Are we done? No! Remember, this is diprotic and so we have excess OH– that will react with HSO3
–. HSO3
– ↔ SO32– + H+ Ka1 = 1.0 x 10–7 pKa2 = 7.000
…continued on next slide….
I 0.59 1.420 — 0.410
C - 0.59 - 0.59 — + 0.59
E 0 0.83 — 1.00
H2SO3 + OH– ↔ H2O + HSO32–
Problem 17 Exam I
A buffer solution based upon sulfurous acid (H2SO3) was created by treating 1.00 L of a 1.00 M sulfurous acid solution with NaOH until a pH of 1.666 was achieved. To this buffer 1.420 moles of NaOH was added. What is the final pH of this solution? (Assume no volume changes) Now, if we have extra OH– (strong base), it will react with the HSO3
– (acid). So let’s construct an ICE table in moles to find out what’s left after reaction: Now use Henderson-Hasselbalch to find the pH : …continued from previous slide….
I 1.00 0.83 — 0
C - 0.83 - 0.83 — + 0.83
E 0.17 0 — 0.83
HSO3– + OH– ↔ H2O + SO3
2–
pH=pKa2
+ log[SO
3
2- ]
[HSO3
- ]
æ
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ö
ø÷
= -log(1.0´10-7)+ log0.83
0.17
æ
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ø÷
pH= 7.689
Careful to use pKa2 since we have added enough NaOH to go past the first equivalence point (and halfway point).
Problem 18 Exam I
Consider the titration of 105 mL of 1.35 M ammonia, NH3 (Kb = 1.8×10-4) with 0.25 M HCl. What is the pH when 283.5 mL of HCl has been added to the ammonia, NH3, solution?
Let’s write out the ICE chart in moles to see what’s left after reaction: Notice this is exactly the halfway point. So we can directly find pH using pKa. Or, in this case, since it’s a base, pOH from pKb and then pH:
I 0.14175 0.0709 0
C - 0.0709 - 0.0709 + 0.0709
E 0.0709 0 0.0709
NH3 + H+ ↔ NH4+
pOH=pKb
= -log(1.8´10-5)
pOH= 4.74
pH= 9.26
Problem 19 Exam I
Rank the equivalence point (EP) of the following titrations from the most acidic to the least acidic in the following order: Most acidic EP < second most acidic EP < third most acidic EP < fourth most acidic EP < least acidic EP Which titration will have the fourth most acidic EP? At equivalence point (EP): nacid = nbase
We know that an acid being titrated will give a basic EP (less acidic EP). This is because at the EP, we no longer have the acid, we just have its conjugate base, which is basic. This means we can narrow our choices to acids being titrated with bases: choices 2 and 4. Between these two, we want the stronger acid (so we can just compare Ka’s) because the weakest acid will give the least acidic EP (i.e. have the strongest conjugate base). 1. 2.0 M NaClO titrated with 1.0 M HCl. 2. 1.0 M NH4Cl titrated with 2.0 M NaOH. Ka (NH4
+) = 5.56×10-10 3. 2.0 M NaHCOO titrated with 1.0 M HCl. 4. 1.0 M HNO2 titrated with 2.0 M NaOH. Ka (HNO2) = 4.00×10-4 5. 2.0 M C5H5N titrated with 1.0 M HCl.
