1. GCHM – Solutions & Electrochemistry
Question Information
Q-Bank
MCAT
Sim
Non-Sim
Subject
General Chemistry
Foundation
GCHM – Solutions & Electrochemistry
Validity
5 years
Author(s)
Reyes, V. M.
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Editor(s)
Passage Media
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Passage
Buffers are very important solutions in chemistry, specially in biochemistry, and the
concept of buffers is critical to the clinical and medical personnel. Just what are
buffers? Buffers are aqueous solutions of a weak acid (or base) and its conjugate
base (or acid). As such, further addition of other acids or bases to the buffer solution
do not alter its pH dramatically. In other words, a buffer is a solution that resists pH
changes, in the cell buffers ensure acid-base homeostasis.
Biochemical reactions in the cell are catalyzed by enzymes which function only
at physiological pH, that is, pH = 7.40 ± It is therefore very crucial for the
viability of life that the cell is able to maintain its pH as closely as possible to this
value in spite of many different reactions taking place simultaneously that generate
as well as consume protons, H+.

2. [A-] pH = pKa + log [HA] The Henderson- Hasselbach Equation:
Consider a weak acid, HA, dissolved in aqueous solution; it dissociates incompletely
as HA  H+ + A- . Since the dissociation is incomplete (i.e., the acid is weak), there
will be a significant quantity of dissolved HA in solution in addition to A- ions. The H+
(proton) will combine with solvent H2O molecules to form H3O+ and will cause the
concentration of H3O+ to be greater than that of OH- and thus make the solution
slightly acidic.
Suppose also that a salt of conjugate base, A-, of the weak acid HA is added to the
above solution, for example, NaA (Na= sodium). This salt will dissolve completely as
NaA  Na+ + A-. The dissolved Na+ will stay as Na+ (as spectator ion) but the A-, being
a weak base, will abstract a proton H+ from water, H2O, resulting in HA (the original
weak acid) and OH-, as in: A- + H2O  HA + OH- . But since A- is a weak base, only a
small proportion of A- anions from NaA will abstract H+ and there will be a significant
quantity of A- in the solution. The OH- released will balance out with the H3O+
formed from the previous reaction.
The Henderson- Hasselbach Equation:
[A-]
pH = pKa + log
[HA]
HA is a weak acid.
A- is its conjugate base.
[ ] refers to molar concentration
Figure 1
The final situation is that there is both HA (a weak acid) and A- (a weak base) in the
same solution, whose concentrations we know with reasonable certainty. With the Ka
of the weak acid HA known in advance, the pH of the buffer can thus be calculated
using the Henderson-Hasselbach equation shown above in Figure 1. (NOTE: Had we
started with a weak base, BOH, dissociating as BOH  B+ + OH-, the derivation of the
pOH of the resulting buffer will be quite similar.)

3. L. Pauling, General Chemistry
Passage References
PMID/Book
Title of Publication or Book
L. Pauling, General Chemistry
(N/A)
en.wikipedia.com, youtube.com
(N/A)
Author’s own lecture notes.
Question Attributes #1
Topic Blueprint
Henderson-Hasselbach Equation
Competency
MCAT: BS-2: Application of Concepts & Principles
To highlight the concept of a weak acid and its conjugate
base, and to apply the Henderson-Hasselbach equation
in calculating the pH of a solution of the same.
Objective
Media ID(s)
Question ID
Question Stem #1
Consider the weak acid, ascorbic acid, which dissociates as shown below with
Ka = 7.9 x Suppose 35.2 gm. of ascorbic acid and 19.8 gm. of sodium ascorbate are dissolved in a liter of pure water. Using the HHE, calculate the pH of the resulting solution. Given: MW of ascorbic acid = ; atomic weight of Na = 22.99

4. The correct ansawer is A, pH = 4.4. The solution goes as follows:
Answer Choices #1 A)
pH = 4.4 B)
pH = 3.8 C)
pH = 3.5 D)
pH = 3.4
Correct: A)
Explanation #1
The correct ansawer is A, pH = The solution goes as follows:
First we need to make the assumption that there were no volume changes upon
dissolution of ascorbic acid and sodium ascorbate into the liter of water.
Then: molar concentration of ascorbic acid = 35.2/ = 0.20 M
MW of sodium ascorbate = MW ascorbic acid – MW hydrogen + MW sodium =
= =
Molar concentration of sodium ascorbate = 19.8/ = 0.10 M
pKa = -log10 (7.9 x 10-5) =
log10 ([ascorbate]/[ascorbic acid]) = log10 (0.2/0.1) =
Thus pH = = 4.4 using the HHE.
(Choice B) This value is incorrect, as the logarithm was reversed, i.e., log10 (0.1/0.2) =
(Choice C) Here, the MW used for sodium ascorbate is (taken directly
from the given value) instead of , thus this value is incorrect
(Choice D) Here, the natural logarithm was used instead of logarithm to base 10,
i.e., ln (0.1/0.2) = = 3.4, and the value is also incorrect
Educational objective: To highlight the concept of a weak acid and its conjugate
base, and to apply the Henderson-Hasselbach equation in calculating the pH of a
solution of the same.

5. 0000000 0000000 References #1 PMID/Book Title of Publication or Book
Verifications #1
Yes / No
The question is at the Application or higher cognitive level.
Yes / No
The question is based on a realistic clinical scenario.
Yes / No
The question has at least one close distracter, and other options have
educational value.
Yes / No
The question is appropriate to the entry level of nursing practice.
Yes / No
The explanation is short and concise, yet thorough.
Yes / No
The question has an appropriate table/flow chart/illustration.