Chemistry 222
Exam 3: Chapters 8-10
80 Points
Name__________________________________________
Fall 2015
Complete five (5) of the following problems. Each problem is worth 16 points. CLEARLY mark
the problem you do not want graded. You must show your work to receive credit for problems
requiring math. Report your answers with the appropriate number of significant figures.
1. Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.180 M
pyridine, C5H5N(aq) (pKb = 8.80) with 0.180 M HBr(aq): (8 points for each part)
a. At the equivalence point
At the equivalence point, all of the weak base has been converted to its conjugate acid.
Therefore, there are (25.0 mL x 0.180 M) = 4.50 mmol of BH+ in the solution, whose total
volume is 25.0 + 25.0 = 50.0 mL (1:1 stoichiometry). This gives a concentration of 0.0900 M
weak base, whose pKa = 14.00 – pKb = 5.20. We can determine the pH using an ICE table:
BH+
= B + H+
I
0.0900
0
0
C
-x
+x
+x
E 0.0900-x
x
x
Ka = [H+][B] =
x2
= 10-5.20 = 6.309x10-6
+
[BH ]
0.0900-x
After some algebra, we find x2+Kax-0.0900Ka = 0 and with the quadratic equation, we find
x = [H+] = 7.50x10-4 M and pH = 3.13
b. After the addition of 22.0 mL of HBr
At this point we have only converted some of the weak base to its conjugate acid. Therefore
we have a buffer solution.
We have added 22.0 mL x 0.180 M = 3.96 mmol of strong acid.
B
+ HBr  BH+ + H2O
Start (mmol) 4.50
3.96
0
0
End (mmol) 0.54
0
3.96
3.96
We can use H-H to determine pH
pH = pK + log mol B = 5.20 + log 0.54 =
mol BH+
3.96
1
4.33
2. For this problem consider the newly discovered kick acid, a triprotic weak acid of the form
H3A.
a. The plot below represents the fraction of dissociation for kick acid. Label the four curves
with their corresponding alphas (H3A, H2A-, HA2-, A3-) and determine the values for
pKa1, pKa2, and pKa3. (6 points)
The pK’s occur where alpha lines
for two protonation states cross at
0.50. Therefore:
pKa1 = 2.75
pKa2 = 4.00
pKa3 = 7.50
H2AH3A
HA2-
A3-
b. Based in your results from part a, use the axes below to sketch the titration curve for the
titration of 20.0 mL of a 0.100 M solution of kick acid with 0.100 M NaOH. Specify the
pH at a minimum of five points on the plot. (10 points)
Given that the formality of our
acid is the same as the NaOH,
we expect equivalence points to
occur at 20, 40, and 60 mL of
titrant added. From our pKa
values, we can estimate the pH
at the first two equivalence
points because the primary
species that are present are the
amphiprotic forms. Midway to
each equivalence point, pH is
determined by pKa’s
At 20 mL, pH = (pKa1+pKa2)/2 = (2.75+4.00)/2 = 3.38
At 40 mL, pH = (4.00 + 7.50)/2 = 5.75
At 10 mL, pH = pKa1 = 2.75
At 30 mL, pH = pKa2 = 4.00
At 50 mL, pH = pKa3 = 7.50
Other useful points would be:
 At 0 mL, which can be treated as a monoprotic weak acid using an ICE table.
Doing so leads to a pH of 1.90.
 At the last equivalence point, we have a solution of A3-, which can be treated as a
monobasic weak base using an ICE table. Doing so leads to a pH of 9.95.
 At an infinite volume of base added, the pH approaches the pH of 0.10 M NaOH,
pH 13.00,
Notice that the first equivalence point is not very sharp. This is also evidenced in the
alpha plots in that H2A- never gets close to 1.
2
3. Answer the following: (8 points each)
a. Why is a buffer solution most effective at a pH within ±1 of the pKa for the acid used to
make the buffer?
The key here is that the buffer capacity for a buffer solution is best when the ratio of A- to HA
is close to 1. This occurs when the pH = pKa. Within ±1 of the pKa the ratio is still
reasonable, between 0.1 and 10. Outside this range, there is too little of either A- or HA to
allow for reasonable buffer capacity because the addition of even a small amount of H+ or
OH- can cause the ratio to change significantly.
b. Why is the pH of a buffer effectively independent of dilution, while the capacity of the
buffer diminishes as it becomes more dilute?
The key points for discussion here are:
• the ratio of weak acid to conjugate base stays essentially constant with dilution
• the buffer capacity suffers as the buffer is diluted because the addition of strong acid or
base will cause a larger shift in the weak acid:conjugate base ratio.
3
4. You need to prepare a pH 4.50 buffer by adjusting the pH of 200.0 mL of a 0.200 M solution
of monosodium oxalate (NaHC2O4). The only acid and base solutions you have available
are 100 mL of 0.100 M NaOH, and 100 mL of 0.100 M HCl. Given these solutions, can you
make your buffer? Justify your answer with appropriate calculations.
(For oxalic acid, pKa1 = 1.25, pKa2 = 4.26.)
There are two equilibria here to consider:
H2C2O4 = H+ + HC2O4HC2O4- = H+ + C2O42-
pKa1 = 1.25
pKa2 = 4.26
Since the desired pH of our buffer is closest to pKa2, we should focus on the C2O42-/HC2O4conjugate acid/base pair for our buffer. This will require the addition of NaOH to convert
some of the HC2O4- to C2O42-. What ratio of C2O42-/HC2O4- is needed?
pH = pKa2 + log mol C2O42mol HC2O44.50 = 4.26 + log mol C2O42mol HC2O4Rearranging shows us that the ratio of mol C2O42-/mol HC2O4- must be 1.738. Do we have
enough NaOH to do this? How many mol NaOH are needed? Initially, we have
200 mL x 0.200 M = 40.0 mmol mol HC2O41.738
=
mol C2O42- =
x
mol HC2O4
40.0 - x
Solving for x tells us that we need 25.39 mmol NaOH to get to our desired ratio. This
requires 25.39 mmol x 1L/0.100mol = 253.9 mL NaOH, but we only have 100, therefore, we
can't make the buffer!
Alternate Approach:
You could also calculate the maximum pH you could make given the solutions you have. If
you add all 100 mL of NaOH to your HA- solution, the you will form:
HA- + OH-  A2- + H2O
Start (mmol) 40.0
10.0
0
0
End (mmol) 30.0
0
10.0
10.0
So, the pH of a solution of 30.0 mmol HA- and 10.0 mmol A2- will be:
pH = pKa2 + log mol A2- = 4.26 + log 10.0 =
30.0
mol HA-
3.78
Since this is the maximum pH we can make with these solutions, and it is less than our
target pH (4.50), we cannot make the buffer with these materials.
4
5. Calculate the pH of the following solutions: (8 points each)
a. A 2.2 x 10-7 F solution of HCl.
Although HCl is a strong acid, the concentration is low enough that we need to consider the
contribution of water as a source of H+. Start by writing mass and charge balance
equations:
Charge Balance: [H+] = [Cl-] + [OH-]
Mass Balance: [Cl-] = 2.2 x 10-7M
Substituting the mass balance expression in to the charge balance and using Kw, we can get
everything in terms of [H+]:
[H+] = 2.2 x 10-7 + Kw/[H+]  [H+]2 – 2.2 x 10-7[H+] - Kw = 0
Solving the quadratic gives [H+] = 2.587 x 10-7 M or pH = 6.59
b. A solution prepared by combining (a) 12.00 mL of 1.00 M NaOH; (b) 20.0 mL of 0.200 M
H2SO4; (c) 5 mL 0.5 M NaNO3; (d) 2.00 mL 1.00 M HCl and (e) 10.00 mL 0.800 M
benzoic acid and diluting to 100.0 mL. The pKa for benzoic acid is 4.20.
This solution contains strong acids and bases, a weak acid and a neutral salt. Since NaNO3
is a neutral salt, is does not impact pH. We first need to determine what is left after the
strong acids and bases react:
NaOH produces: 10.00 mL x 1.00 M NaOH x 1 mol OH-/1 mol NaOH = 12.00 mmol OHH2SO4 produces: 20.0 mL x 0.200 M H2SO4 x 2 mol H+/1 mol H2SO4= 8.00 mmol H+
HCl produces: 2.00 mL x 1.00 M HCl x 1 mol H+/1 mol HCl = 2.00 mmol H+
Therefore, our strong acids produce 8.00 + 2.00 = 10.00 mmol H+ and our strong base will
produce 12.00 mmol OH-. After reaction, 2.00 mmol of OH- remains, which can further react
with our weak acid. The result is a buffer solution that contains both HA and A--.
HA
+
OH→
A+ H2O
Start 8.00 mmol
2.00 mmol
0
End 6.00 mmol
0
2.00 mmol
Since this is a buffer solution, we can use the H-H equation to determine pH:
pH = pKa + log mol A- = 4.20 + log 2.00 =
mol HA
6.00
(An ICE table approach would lead to the same pH.)
5
3.72
6. Calculate the pH and concentrations (in moles per liter) of H2A, HA- and A2- in a solution
prepared by dissolving 0.200 moles of NaHA in 1.00 L of solution.
(For H2A: Ka1 = 1.0x10-3, Ka2 = 1.0x10-7)
First find the [H+]:
H  

K a1K a 2F  K a1K w

K a1  F
10 10 0.200M  10 10   9.98 x10
3
7
3
 14
10  3  0.200M
6
M
pH = -log(9.98x10-6) = 5.00
(You get the same result if you average the two pKa’s, ½(3.00 + 7.00) = 5.00)
Once we have [H+], we can find the concentrations by determining the fraction of each
species present. [X] = xF.
 H2 A 
H 
H   H  K
 2
 2

a1
 K a1K a 2
In all cases, the denominator of the alpha expression is the same:
d = [H+]2 + Ka1[H+] + Ka1Ka2 = (9.98x10-6)2 + (10-3)(9.98x10-6) + (10-3)(10-7) = 1.017x10-8
For H2A:
+ 2
-6 2
H2A = [H ] = (9.98x10 ) = 0.00978
d
1.017x10-8
and [H2A] = H2AF = 0.00978x0.200M = 0.00196M
For HA-:
+
-3
-6
HA- = Ka1[H ] = (10 )(9.98x10 ) = 0.980
d
1.017x10-8
and [HA-] = HA-F = 0.980x0.200M = 0.196M
For A2-:
-3
-7
A2- = Ka1Ka2 = (10 )(10 ) = 0.00983
d
1.017x10-8
and [A2-] = A2-F = 0.00983x0.200M = 0.00197M
6
Possibly Useful Information
H  

K a1K a 2F  K a1K w
 K a1K a 2
K a1  F
pH 
conjugate base 
pH  pK a  log
weak acid 
 A 2 
 H2 A
H 

H   H  K
 2
 2
H   H  K

a1

a1
K a1K a 2
 2
1
pK a1  pK a 2 
2
 K a1K a2
K aK b = K w
 K a1K a 2
Kw = 1.0 x 1014 = [H+][OH-]
x
7
 b  b 2  4ac
2a