• Wrong document? Swap it for free
  • Written by students who passed
  • Immediately available after payment
  • Read online or as PDF
Sell
Where do you study
Your language
Document preview thumbnail
Preview 4 out of 84 pages
Exam (elaborations)

Chem 210 Biochemistry Exam With Verified Answers Detailed Rationales Graded A+

Document preview thumbnail
Preview 4 out of 84 pages

Chem 210 Biochemistry Exam With Verified Answers Detailed Rationales Graded A+

Content preview

Page 1 of 84


CHEM 210 BIOCHEMISTRY EXAM WITH VERIFIED
ANSWERS DETAILED RATIONALES GRADED A+




CHEM 210 Biochemistry — Module 1 to 8 Exams

SECTION 1: WATER, pH & BUFFERS (Questions 1–30)

1. Which property of water allows it to dissolve ionic compounds?

A. Low dielectric constant
B. High dielectric constant and ability to form hydrogen bonds
C. Low specific heat
D. Nonpolar nature

Rationale: Water has a high dielectric constant, which reduces the electrostatic attraction
between ions, and it forms hydrogen bonds with ions, stabilizing them in solution. This makes
water an excellent solvent for ionic compounds.

2. The pH of a solution with [H⁺] = 1 × 10⁻⁴ M is:

A. 4
B. 10
C. 1
D. 14

Rationale: pH = −log[H⁺] = −log(1 × 10⁻⁴) = 4. The pH scale is logarithmic and inversely related to
hydrogen ion concentration.


1|Page

,Page 2 of 84


3. What is the pH of a 0.01 M solution of HCl?

A. 1
B. 2
C. 12
D. 0.01

Rationale: HCl is a strong acid that completely dissociates. [H⁺] = 0.01 M = 1 × 10⁻² M. pH =
−log(10⁻²) = 2.

4. The Henderson-Hasselbalch equation is:

A. pH = pKa + log([A⁻]/[HA])
B. pH = pKa − log([A⁻]/[HA])
C. pH = pKa + log([HA]/[A⁻])
D. pH = pKa × log([A⁻]/[HA])

Rationale: The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]), where [A⁻] is the
concentration of the conjugate base and [HA] is the concentration of the weak acid.

5. At a pH equal to the pKa of a weak acid:

A. The acid is completely protonated
B. The acid is completely deprotonated
C. [HA] = [A⁻]
D. pH = 7

Rationale: When pH = pKa, the concentrations of the weak acid (HA) and its conjugate base (A⁻)
are equal. This is the point of maximum buffering capacity.

6. Which of the following is a biological buffer system?

A. HCl/NaOH
B. H₂CO₃/HCO₃⁻
C. NaCl/H₂O
D. Glucose/fructose

2|Page

,Page 3 of 84


Rationale: The carbonic acid-bicarbonate buffer system (H₂CO₃/HCO₃⁻) is a major biological
buffer that helps maintain blood pH within a narrow range.

7. What is the pH of pure water at 25°C?

A. 0
B. 7
C. 14
D. 1

Rationale: Pure water has a pH of 7 at 25°C, which is considered neutral because [H⁺] = [OH⁻] =
1 × 10⁻⁷ M.

8. The pKa of acetic acid is 4.76. What is the pH of a buffer containing 0.1 M acetic acid and
0.1 M sodium acetate?

A. 4.76
B. 7.00
C. 5.76
D. 3.76

Rationale: Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) = 4.76 +
log(0.1/0.1) = 4.76 + 0 = 4.76.

9. Which of the following statements about water is TRUE?

A. Water is a nonpolar molecule
B. Water has a low boiling point compared to other molecules of similar size
C. Water has a high specific heat capacity
D. Water cannot form hydrogen bonds

Rationale: Water has a high specific heat capacity (4.18 J/g·°C), which allows it to absorb
significant heat with minimal temperature change, helping to regulate body temperature.

10. A buffer is most effective when:


3|Page

, Page 4 of 84


A. pH is far from pKa
B. pH = pKa
C. pH is 2 units above pKa
D. pH is 2 units below pKa

Rationale: A buffer is most effective when the pH is within ±1 pH unit of the pKa, with maximum
buffering capacity at pH = pKa.

11. What is the conjugate base of H₂PO₄⁻?

A. H₃PO₄
B. HPO₄²⁻
C. PO₄³⁻
D. H₂O

Rationale: The conjugate base is formed by removing a proton (H⁺) from the acid. H₂PO₄⁻ →
HPO₄²⁻ + H⁺, so HPO₄²⁻ is the conjugate base.

12. The pH of a solution with [OH⁻] = 1 × 10⁻⁵ M is:

A. 5
B. 9
C. 14
D. 1

Rationale: pOH = −log(10⁻⁵) = 5. pH = 14 − pOH = 14 − 5 = 9.

13. Which of the following is NOT a property of water?

A. High surface tension
B. High dielectric constant
C. Low heat of vaporization
D. High specific heat

Rationale: Water has a high heat of vaporization (40.7 kJ/mol), not a low one, which is
important for sweating and evaporative cooling.

4|Page

Document information

Uploaded on
October 10, 2026
Number of pages
84
Written in
2026/2027
Type
Exam (elaborations)
Contains
Questions & answers
$26.99

Wrong document? Swap it for free Within 14 days of purchase and before downloading, you can choose a different document. You can simply spend the amount again.
Written by students who passed
Immediately available after payment
Read online or as PDF

Sold
3
Followers
0
Items
594
Last sold
2 weeks ago



Why students choose Stuvia

Created by fellow students, verified by reviews

Quality you can trust: written by students who passed their tests and reviewed by others who've used these notes.

Didn't get what you expected? Choose another document

No worries! You can instantly pick a different document that better fits what you're looking for.

Pay as you like, start learning right away

No subscription, no commitments. Pay the way you're used to via credit card and download your PDF document instantly.

Student with book image

“Bought, downloaded, and aced it. It really can be that simple.”

Alisha Student

Working on your references?

Create accurate citations in APA, MLA and Harvard with our free citation generator.

Working on your references?

Frequently asked questions