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. Reviewer(s) 0000000 Editor(s) 0000000 Passage Media Media ID(s) 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 ± .02. 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+.

[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.)

L. Pauling, General Chemistry Passage References PMID/Book Title of Publication or Book 000000000 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) 00000000 Question ID 00000000 Question Stem #1 Consider the weak acid, ascorbic acid, which dissociates as shown below with Ka = 7.9 x 10-5. 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 = 176.12; atomic weight of Na = 22.99

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 = 4.4. 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 = 35.2/176.12 = 0.20 M MW of sodium ascorbate = MW ascorbic acid – MW hydrogen + MW sodium = 176.12 -1.00 + 22.99 = 198.11 = 198.11 Molar concentration of sodium ascorbate = 19.8/198.11 = 0.10 M pKa = -log10 (7.9 x 10-5) = 4.1024 log10 ([ascorbate]/[ascorbic acid]) = log10 (0.2/0.1) = 0.3103 Thus pH = 4.1024 + 0.3103 = 4.4 using the HHE. (Choice B) This value is incorrect, as the logarithm was reversed, i.e., log10 (0.1/0.2) = -0.3103 (Choice C) Here, the MW used for sodium ascorbate is 22.99 (taken directly from the given value) instead of 198.11, 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) = -0.693 = 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.

0000000 0000000 References #1 PMID/Book Title of Publication or Book 000000000 000000000 0000000 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.