To get class handouts for Buffers and Titrations click HERE.
Here are some typical questions concerning buffers and titrations.
| 1. Which of the following would
be to form a buffer to control a pH of 10.0? |
| a. HC2H3O2 -
NaC2H3O2 |
d. NaHCO3 - Na2CO3 |
| b. NaH2PO4 - Na2HPO4
|
e. HF - NaF |
| c. H3PO4 - NaH2PO4 |
| 2. 400.0 mL of 0.15 M KOH was mixed with 450
mL of 0.20 M acetic acid, HOAc. What is the pH of the final solution?
Ka (HOAc) = 1.7 x 10-5 pH = 5.07 |
| 3. Consider a buffer
composed of 0.450 M propionic acid, HC3H5O2(aq) and 0.350
M sodium propionate, NaC3H5O2(aq). Ka (HC3H5O2)
= 1.3 x 10-5 |
| a. Write the equation
for the chemical equilibrium that exists in this buffer: HC3H5O2(aq)
<=====> C3H5O2-(aq) + H+(aq) |
| b. What is the pH of
this buffer? pH = 4.78 |
| c. What is the pH of
this buffer after 0.0180 mol of HCl(aq) is added to 1.0 L of it?
pH = 4.74 |
| d. In words, with
equations as needed, describe how this buffer operates to control the pH when HCl(aq)
is added to it. Be specific. |
|
i.
System is at equilibrium. ii. Acid increases hydronium ion concentration.
iii. Equilibrium shifts to the left
to offset the increase.
iv. The propionic acid conc
increases, while propionate decreases.
v. New equilibrium is
reestablished with the pH a little lower than before. |
4. How many grams of sodium acetate, NaC2H3O2,
must be added to 2.000 L of 0.500 M acetic acid to prepare a buffer with a pH of
5.00? (for HC2H3O2: Ka = 1.8 x 10!5,
pKa = 4.74, FW = 60.0; and for NaC2H3O2:
FW = 82.0) Assume no volume change with the addition of sodium acetate.
answer = 149-150 g |
5. What is the pH of the buffer composed of 0.750 mol of
aqueous ammonia, NH3(aq), and 0.900 mol of ammonium bromide, NH4Br?
(for NH3(aq): Kb = 1.8 x 10-5 )
pH = 9.17
|
| 6. In terms of the molecular structure of two
acids, why is the Ka of chlorous acid, HClO2, 1.1 x 10-2,
a substantially larger value than that for hypochlorous acid, HClO, which is
3.5 x 10-8? Both Ka's are in water at 25oC.
Again, be specific. The more
polar is the OH bond, the stronger the acid in water. The added inductive effect of the
second oxygen on the chlorine atom in chlorous acid makes the OH bond more polar than that
in HClO. |
| 7. (a) What is the pH of a buffer composed of
0.500 mole benzoic acid, HC7H5O2, Ka =
6.3 x 10-5, and 0.350 mole sodium benzoate, NaC7H5O2
in 1.500 L of solution? pH = 4.05 |
(b) What is the pH of this buffer after 0.075 mol of HCl is
added to it?
pH = 3.88 |
(c) What would be the pH change in 1.500 L of pure
water (pH = 7.00) if 0.075 mol of HCl is added to it?
pH change =
decreases 5.7 units |
| (d) Explain how the buffer controlled the
change in pH better than pure water when 0.075 mole of HCl was added to it. The buffer contained a base (benzoate ion) to
consume the added acid as it is converted to the weak acid benzoic acid. This
reduces the hydronium ion concentration in solution to a level just a little higher than
it was before the addition of the HCl. Pure water does not have the ability to
neutralize acid. |
8. What must the mole ratio of sodium benzoate to benzoic
acid to prepare a buffer with a pH of 3.80? Ka for benzoic acid is 6.3 x 10-5.
[NaC7H5O2]/
[HC7H5O2] = 0.40
|
9. In a titration, how many milliliters of 0.1105 M NaOH
would be required to just neutralize all the acid in 25.00 mL of 0.09950 M nitric acid?
volume NaOH = 22.51
mL |
| Which would be the best
choice for an indicator for this titration? Circle one. |
| methyl yellow pKa = 3.3 |
bromophenol blue pKa = 3.85 |
| methyl red pKa = 4.95 |
natural red pKa =
7.4 |
| cresol red pKa = 8.2 |
thymolphthalein pKa = 10.0 |
| 10. A titration curve is shown below for the
titration of 20.00 mL 0.1500 M weak acid, HA, with 0.1132 M sodium hydroxide. Ka(HA)
= 3.0 x 10-7. You are to calculate the pH at points A, B, C
and D. |
 |  |
| Again, Ka(HA) = 3.0 x 10-7. |
| (a) What is the pH at point A, before any
NaOH is added to the acid? pH = 3.67 |
(b) What is the pH at point B, the point where exactly half
the acid is neutralized?
pH = 6.52 |
(c) What is the pH at point C, where exactly three-fourths
of the acid is neutralized?
pH = 7.00 |
(d) What is the pH at point D, the equivalence point of the
titration?
concentration of salt = 0.0645 M
and Kb = 3.3 x 10-8
pH = 9.67
|
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