Problem #1 - What is the pH of each of the following solutions?
a). 0.04
M HCl
Strong acids completely dissociate in solution therefore the total concentration is equal to the concentration of H+. pH = -log[H+] = -log [ 0.04 M ] = 1.4

Problem #1 - What is the pH of each of the following solutions?
b). 0.04
M NaH
2
PO
4
The dissociation of weak acids in solution is not complete so you must first calculate the concentration of the H+ using the Ka for the weak acid.
Ka for NaH
2
PO
4
= 6.23 x 10 -8 M
[H+] 2 [H+] 2
Ka = ––––– = –––– = 6.23 x 10 -8 M
[HA] [ 0.04
]
[H+] = √(Ka)(Total concentration)
= √( 6.23 x 10 -8 M )( 0.04
) = 5.0 x 10 -5 M pH = -log [5 x 10 -5 M] = 4.3

Problem #1 - What is the pH of each of the following solutions?
c). 0.1
M H
2
CO
3
(HA) and 0.6
M NaHCO
3
(A ) pH = pKa + log [A-]
[HA] pH = 6.4 + log ( 0.6
/ 0.1) = 7.2
( H
2
CO
3 pKa=6.4)

Problem #1 - What is the pH of each of the following solutions?
d). 0.1
M H
2
CO
3
(HA) and 0.08
M NaOH (acceptor) pH = pKa + log [A-]
[HA] pH = 6.4 + log 0.08
/( 0.1
– 0.08
) = 7.0
(NaOH will completely dissociate)
(H
2
CO
3 pKa=6.4)
Here you are figuring out to find out how much of the buffer was used up by the base (NaOH).

Problem #1 - What is the pH of each of the following solutions?
e). 0.45
M NaHCO
3
(A )and 0.35
M HCl (donor) pH = pKa + log [A-]/[HA] =
6.4 + log ( 0.45
– 0.35
)/ 0.35
= 5.9
(HCl will completely dissociate)
One of the forms (not balanced): NaHCO
3
So use pKa of H
2
CO
3
(6.4)
->H
2
CO
3
HA + OH -> A + H
2
O
A + H + -> HA
Here you are figuring out to find out how much of the buffer was used up by the acid (HCl).

Problem #2 - Determine the H+ and OHconcentration in cow’s milk. The pH of cow’s milk is typically 6.5
.
The [H+] can be determined from the equation pH = -log [H+].
6.5
= - log [H+] or
10 6.5
= [H+] = 3.2 x 10 -7 M
Since the ion product of water is
Kw= [H+][OH-]
1 x 10 14 =[ H+ ][OH-]
[OH-] = 1 x 10 14 / 3.2 x 10 7 = 3.2 x 10 -8 M.

Problem #3 - Sketch a titration curve for tyrosine.
What is the isoelectric point of tyrosine? Use table 3.2 for pKa values.
α
-amino (9.2)
α
-carboxyl (2.2)
Increasing [OH-]
R group (10.5)

To determine the pI of tyrosine first calculate the net charge for tyrosine at pH 2.2, 9.2 and 10.5
(these values are equal to the pKa values for the three ionizable hydrogens of tyrosine).
α-amino (9.2) R group (10.5) α-carboxyl (2.2) Net Charge pH 2.2
pH<< pKa
100%HA
All NH3+ pH<< pKa
100%HA
All OH pH = pKa
50% HA/50% A-
½ COOH+½ COO-
0.5
+ 1 pH 9.2
pH = pKa 50%
HA/50% A-
½ NH3 + ½ NH2
0 pH<< pKa
100%HA
All OH
0.5
pH
 pKa
100%A-
All COO-
0.5
pH
10.5
+ 0.5
pH
 pKa 100%A-
All NH2
0
+ 0 pH = pKa 50%
HA/50% A-
½ OH ½ O-
0.5
1 pH
 pKa
100%A-
All COO-
1
1.5
The pI is the pH at which the net charge on the molecule is 0 which in this case lands between pH 2.2 and 9.2. The average of these two values is
(2.2 + 9.2)/2 = 11.4/2 = 5.7

Problem #4 – How many moles of NaOH would you need to add to 800 ml of a 0.4 M
H
2
CO
3 solution in order to bring the pH of the solution to 10.5?
We have 0.32
moles of H
2
CO
3
(800 ml of 0.4 M) and we need to figure out how much NaOH needs to be added to get a final pH of 10.5
When you add 0.32 moles of (equal amount) you convert all the H
2
CO
3
Next, find the ratio of HCO
3
[HA] and CO
3
-to HCO
3
-
[A ] needed to achieve a pH of 10.5: pKa HCO
3
= 10.2
10.5
= 10.2
+ log[A ]/[HA]
0.3 = log[A ]/[HA]
10^(0.3) = 10^(log[A ]/[HA])  2 = [A ]/[HA]
0.32 = [HA] + [A ] or [A ] = 0.32
- [HA]
2 = [A ]/[HA]
So 2 = (0.32 - [HA])/[HA]  [HA] = 0.103
0.32 = 0.103 + [A ]
[A ] = 0.32 - 0.103
[A ] = 0.21 moles
We need to add another 0.21
moles of NaOH to our starting amount ( 0.32 mol)
So 0.21
moles + 0.32 moles = 0.53
moles will yield a pH of 10.5.

Problem #5 a)What is the pH of a 0.6 M solution of benzoic acid ( Ka 6.5 x 10 -4 )?
[H+] 2 [H+] 2
Ka = ––––– = –––– = 6.5 x 10 -4 M
[HA] [ 0.6
]
[H+] = √(Ka)(Total concentration)
= √( 6.5 x 10 -4 M )( 0.6
) = 0.0197 M pH = -log [0.0197 M] = 1.7
b) What is the pH of a solution containing 0.03M benzoic acid (p Ka 3.19) and 0.02M Nabenzoate?
pH = pKa + log [A-]
[HA] pH = 3.19 + log ( 0.02
/ 0.03) = 3.0

Problem #6
Sally, your 199 supervisor, asks for a pH 7.5 buffer. You already have 0.5M
KH
2
PO
4
. What concentration of K
2
HPO
4 do you need? (p K a = 6.86) pH = pKa + log [A-]
[HA]
7.5 = 6.86+ log ([K
2
HPO
4
]/0.5)
0.64 = log ([K
2
HPO
4
]/0.5)
4.365 = [K
2
HPO
4
]/0.5
2.18 M = [K
2
HPO
4
]

Problem #7 True or False a) One of the reasons water is a polar molecule is because it is V-shaped
(rather than linear). b) Buffers are aqueous systems that tend to resist changes in pH when small amounts of H+ and OH- are added. c) The [H+] is 1M in a solution at pH 4.0. d) Adding a strong base to a weak acid solution will shift HA molecules to the A- form. e) A single water molecule can act as a hydrogen bond donor in as many as four hydrogen bonds at the same time.
f) An acid that dissociates to the extent of 92% in water would be termed a strong acid. g) Lactic acid (pKa 3.9) is a stronger acid than phosphoric acid (pKa 2.1). h) The pKa is the pH at which a weak acid’s protons are 50% titrated. i)The oxygen atom in water has a partial positive charge.
j) Hydrogen bonds are relatively weak compared to covalent bonds.
T
T
F
T
F
F
F
T
F
T

If you have enough time!
Problem #8 - When you make salad dressing the oil and vinegar always separate.
Why?
The oil is composed of long chain nonpolar fatty acid molecules while the vinegar is a water solution of dilute acetic acid and so contains mostly water molecules. The fat molecules are not soluble in water owing to the hydrophobic effect. In order to solvate the fat molecules the water must become highly ordered around the fat molecules which is thermodynamically unfavorable. This has a very unfavorable
D
S. When the fat molecules come together and squeeze out the water, the water can now adopt the much more favorable hydrogen bonding pattern unique to water.
So overall it is much more energetically/thermodynamically for the fat molecules to interact with one another than water. Hence, water and oil separate. The oil goes to the top because it is less dense than water.