# Worksheet1_Solutions

```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 NaH2PO4
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 NaH2PO4= 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 H2CO3 (HA) and 0.6 M NaHCO3 (A-)
pH = pKa + log [A-]
[HA]
pH = 6.4 + log (0.6/0.1) = 7.2
(H2CO3 pKa=6.4)
Problem #1 - What is the pH of each of the following
solutions?
d). 0.1 M H2CO3 (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)
(H2CO3 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 NaHCO3 (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): NaHCO3 ->H2CO3
So use pKa of H2CO3 (6.4)
HA + OH- -> A + H2O
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 10-7M
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 10-8M.
Problem #3 - Sketch a titration curve for tyrosine.
What is the isoelectric point of tyrosine? Use
table 3.2 for pKa values.
Increasing pH
R group (10.5)
α-amino (9.2)
α-carboxyl (2.2)
Increasing [OH-]
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).
pH 2.2
pH 9.2
pH
10.5
α-amino (9.2)
R group (10.5)
α-carboxyl (2.2)
Net Charge
pH<< pKa
100%HA
All NH3+
pH<< pKa
100%HA
All OH
pH = pKa
50% HA/50% A½ COOH+½ COO-
0.5
+1
0
-0.5
pH = pKa 50%
HA/50% A½ NH3 + ½ NH2
pH<< pKa
100%HA
All OH
pH pKa
100%AAll COO-
+0.5
+0
-1
pH pKa 100%AAll NH2
pH = pKa 50%
HA/50% A½ OH ½ O-0.5
pH pKa
100%AAll COO-1
0
-0.5
-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
H2CO3 solution in order to bring the pH of the solution to 10.5?
We have 0.32 moles of H2CO3 (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 H2CO3 to HCO3Next, find the ratio of HCO3- [HA] and CO3-- [A-] needed to achieve a pH of 10.5:
pKa HCO3- = 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 (pKa 3.19)
and 0.02M Nabenzoate?
pH = pKa + log [A-]
[HA]
pH = 3.19 + log (0.02/0.03) = 3.0
Problem #6
KH2PO4. What concentration of K2HPO4 do you need? (pKa = 6.86)
pH = pKa + log [A-]
[HA]
7.5 = 6.86+ log ([K2HPO4]/0.5)
0.64 = log ([K2HPO4]/0.5)
4.365 = [K2HPO4]/0.5
2.18 M = [K2HPO4]
Problem #7
True or False
a) One of the reasons water is a polar molecule is because it is V-shaped
(rather than linear).
T
b) Buffers are aqueous systems that tend to resist changes in pH when
small amounts of H+ and OH- are added.
T
c) The [H+] is 1M in a solution at pH 4.0.
F
d) Adding a strong base to a weak acid solution will shift HA molecules to
the A- form.
T
e) A single water molecule can act as a hydrogen bond donor in as many
as four hydrogen bonds at the same time.
F
f) An acid that dissociates to the extent of 92% in water would be termed a
strong acid.
F
g) Lactic acid (pKa 3.9) is a stronger acid than phosphoric acid (pKa 2.1).
F
h) The pKa is the pH at which a weak acid’s protons are 50% titrated.
T
i)The oxygen atom in water has a partial positive charge.
F
j) Hydrogen bonds are relatively weak compared to covalent bonds.
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
DS. 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.
```
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