CHEM134 C002 Spring 2026
Smart NCLEX-Style Chemistry Practice Exam
Category: CHEM134 C002 Spring 2026
Topic: Digital Titration of Acids and Bases + General Chemistry II Review
Format: Scenario-based multiple-choice questions with answer, rationale, key terms, and answer-choice analysis.
Academic note: This is an original practice exam generated for study and revision. It is not an official CHEM134
exam and is not copied from a paid exam bank.
Suggested use: Try each question first, then review the rationale and the incorrect-answer explanations. For
calculation items, write the equation before choosing an answer.
Answer Key
Q Answer Topic
1 A Acid-base titration curve interpretation
2 A Titration apparatus and measurement error
3 B Monoprotic acid concentration from titration data
4 A pH before equivalence in strong acid-strong base titration
5 B Weak acid-strong base equivalence point
6 A Buffer pH calculation
7 B Buffer selection
8 C Solubility equilibrium and Ksp
9 D Le Chatelier principle
10 B Rate law from initial rates
11 C First-order half-life
12 A Catalysts and activation energy
13 B Gibbs free energy and spontaneity
14 C Galvanic cells
15 A Nuclear chemistry
16 C Colligative properties
17 B Dilution calculation
18 A Systematic error in titration
19 A Beer-Lambert law
20 A Indicator choice
21 B Laboratory safety and response
CHEM134 C002 Spring 2026 - Smart NCLEX-Style Practice Exam Page 1
, Question 1
Category: CHEM134 C002 Spring 2026
Topic: Acid-base titration curve interpretation
Stem: Diagram 1 shows 25.00 mL of 0.100 M acetic acid titrated with 0.100 M NaOH. At point H, what conclusion
should the student make?
A. The solution is at half-equivalence; pH = pKa and [CH3CO2H] = [CH3CO2-].
B. The solution is at equivalence; all acid has been converted to conjugate base.
C. The solution is beyond equivalence; excess NaOH controls the pH.
D. The solution is a strong acid-strong base mixture with pH = 7.
Correct Answer: A
Rationale: At half-equivalence, exactly half of the weak acid has been neutralized to conjugate base. The
Henderson-Hasselbalch equation becomes pH = pKa because log([base]/[acid]) = log(1) = 0.
Key terms explained: Half-equivalence point: the point where moles of base added equal half the original moles of
weak acid. pKa: a measure of weak acid strength. Buffer: a mixture that resists pH change because it contains a
weak acid and its conjugate base.
Why the others are not correct:
A. Correct - this choice matches the chemistry principle or calculation.
B. Equivalence occurs at 25.00 mL, not 12.50 mL.
C. Excess NaOH controls pH only after the equivalence point.
D. Acetic acid is weak, so this is not a strong acid-strong base titration.
Question 2
Category: CHEM134 C002 Spring 2026
Topic: Titration apparatus and measurement error
CHEM134 C002 Spring 2026 - Smart NCLEX-Style Practice Exam Page 2
Smart NCLEX-Style Chemistry Practice Exam
Category: CHEM134 C002 Spring 2026
Topic: Digital Titration of Acids and Bases + General Chemistry II Review
Format: Scenario-based multiple-choice questions with answer, rationale, key terms, and answer-choice analysis.
Academic note: This is an original practice exam generated for study and revision. It is not an official CHEM134
exam and is not copied from a paid exam bank.
Suggested use: Try each question first, then review the rationale and the incorrect-answer explanations. For
calculation items, write the equation before choosing an answer.
Answer Key
Q Answer Topic
1 A Acid-base titration curve interpretation
2 A Titration apparatus and measurement error
3 B Monoprotic acid concentration from titration data
4 A pH before equivalence in strong acid-strong base titration
5 B Weak acid-strong base equivalence point
6 A Buffer pH calculation
7 B Buffer selection
8 C Solubility equilibrium and Ksp
9 D Le Chatelier principle
10 B Rate law from initial rates
11 C First-order half-life
12 A Catalysts and activation energy
13 B Gibbs free energy and spontaneity
14 C Galvanic cells
15 A Nuclear chemistry
16 C Colligative properties
17 B Dilution calculation
18 A Systematic error in titration
19 A Beer-Lambert law
20 A Indicator choice
21 B Laboratory safety and response
CHEM134 C002 Spring 2026 - Smart NCLEX-Style Practice Exam Page 1
, Question 1
Category: CHEM134 C002 Spring 2026
Topic: Acid-base titration curve interpretation
Stem: Diagram 1 shows 25.00 mL of 0.100 M acetic acid titrated with 0.100 M NaOH. At point H, what conclusion
should the student make?
A. The solution is at half-equivalence; pH = pKa and [CH3CO2H] = [CH3CO2-].
B. The solution is at equivalence; all acid has been converted to conjugate base.
C. The solution is beyond equivalence; excess NaOH controls the pH.
D. The solution is a strong acid-strong base mixture with pH = 7.
Correct Answer: A
Rationale: At half-equivalence, exactly half of the weak acid has been neutralized to conjugate base. The
Henderson-Hasselbalch equation becomes pH = pKa because log([base]/[acid]) = log(1) = 0.
Key terms explained: Half-equivalence point: the point where moles of base added equal half the original moles of
weak acid. pKa: a measure of weak acid strength. Buffer: a mixture that resists pH change because it contains a
weak acid and its conjugate base.
Why the others are not correct:
A. Correct - this choice matches the chemistry principle or calculation.
B. Equivalence occurs at 25.00 mL, not 12.50 mL.
C. Excess NaOH controls pH only after the equivalence point.
D. Acetic acid is weak, so this is not a strong acid-strong base titration.
Question 2
Category: CHEM134 C002 Spring 2026
Topic: Titration apparatus and measurement error
CHEM134 C002 Spring 2026 - Smart NCLEX-Style Practice Exam Page 2
