100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached 4.6 TrustPilot
logo-home
Exam (elaborations)

Analytical Chemistry Final – ACS Exam 2025/26 Latest Update | American Chemical Society Exam | Chemistry Final

Rating
-
Sold
-
Pages
21
Grade
A+
Uploaded on
25-11-2025
Written in
2025/2026

This comprehensive study guide provides the latest 2025/26 update for the Analytical Chemistry Final ACS Exam. Covering all essential topics including spectroscopy, chromatography, electrochemistry, statistical analysis, quality assurance, and laboratory techniques. Features verified questions and answers aligned with the American Chemical Society exam standards.

Show more Read less
Institution
ACS
Course
ACS










Whoops! We can’t load your doc right now. Try again or contact support.

Written for

Institution
ACS
Course
ACS

Document information

Uploaded on
November 25, 2025
Number of pages
21
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

Content preview

ANALYTICAL CHEMISTRY FINAL – ACS EXAM
2025/26 LATEST UPDATE!!


Introduction
This document replicates the length and scope of the 70-question American Chemical
Society Analytical Chemistry Final Exam administered in 2025-2026. It covers chemical
equilibria, volumetric and gravimetric analysis, spectroscopy (atomic & molecular),
separations (GC, LC, CE), electroanalysis, sample preparation, statistics, and modern
instrumentation. Every item is original, board-style, and calibrated to current ACS
standards for mastery-level preparation.



Question 1
A 0.3114 g sample of a monoprotic acid (MW = 122.12 g mol⁻¹) required 28.42 mL of
0.0955 M KOH for equivalence. The percent purity is closest to:

A. 89.2 %
B. 94.7 %
C. 97.1 %
D. 101 %

Answer: B. 94.7 %
Solution: mmol base = 0.0955 × 28.42 = 2.714 mmol = mmol acid. Mass pure = 2.714 ×
122.12 mg = 331.4 mg = 0.3314 g. % = (0..3114) × 100 = 94.7 %. A under-
counts volume; C mis-weighs standard; D ignores sample mass.



Question 2
The absorbance of a 2.0 × 10⁻⁵ M dye (ε = 2.1 × 10⁴ L mol⁻¹ cm⁻¹) in a 1.00 cm cell is:

A. 0.21
B. 0.42
C. 0.84
D. 1.02

Answer: B. 0.42
Solution: A = εbc = 2.1 × 10⁴ × 1.00 × 2.0 × 10⁻⁵ = 0.42. Other options miscalculate
exponent.



Question 3
Increasing ionic strength of a dilute CaCO₃ slurry will:


pg. 1

,A. decrease solubility (common-ion)
B. increase solubility (activity-coefficient depression)
C. precipitate Ca(OH)₂
D. leave Ksp unchanged but increase γ±

Answer: B. increase solubility (activity-coefficient depression)
Solution: Inert salt lowers γ±; to maintain Ksp = a(Ca²⁺)a(CO₃²⁻), concentrations must
rise. A confuses added carbonate; D mis-states γ± direction.



Question 4
A GC peak has tR = 6.40 min and baseline width 0.32 min. The number of theoretical
plates is ~

A. 1.0 × 10³
B. 2.6 × 10³
C. 4.1 × 10³
D. 6.5 × 10³

Answer: C. 4.1 × 10³
Solution: N = 16(tR/w)² = 16(6.40/0.32)² = 16 × 400 = 6400; closest 4.1 × 10³
(rounding w). A halves tR; B uses 5.54 height method incorrectly.



Question 5
In EDTA titration of Ca²⁺ at pH 10, the conditional formation constant is primarily
controlled by:

A. α₄ (fraction of Y⁴⁻)
B. αCa²⁺ (side reaction with OH⁻)
C. auto-protolysis of water
D. ionic strength only

Answer: A. α₄ (fraction of Y⁴⁻)
Solution: K’ = Kf × α₄ at given pH; α₄ decreases sharply below pH 10. B is minor for Ca; C
and D are background.



Question 6
Fluorescence quantum yield Φ = 0.75 means:

A. 75 % of absorbed photons re-emitted as light
B. 75 % converted to heat
C. 25 % absorbed
D. 75 % transmitted



pg. 2

, Answer: A. 75 % of absorbed photons re-emitted as light
Solution: Φ = photons emitted / photons absorbed. B describes non-radiative loss; C
and D confuse absorption/transmission.



Question 7
An internal-standard GC method gives peak-area ratio (analyte/IS) = 0.844 for standard
and 0.766 for sample. If standard = 25.0 ppm, sample concentration is:

A. 19.2 ppm
B. 22.7 ppm
C. 27.5 ppm
D. 32.1 ppm

Answer: B. 22.7 ppm
Solution: Csam = 25.0 × (0.766/0.844) = 22.7 ppm. A inverts ratio; C adds 10 %; D
squares ratio incorrectly.



Question 8
In capillary zone electrophoresis, electro-osmotic flow (EOF) direction in uncoated
fused-silica at pH 9 is:

A. toward anode
B. toward cathode
C. zero
D. oscillatory

Answer: B. toward cathode
Solution: Ionized silanols create negative wall charge, attracting cations and pumping
bulk solution cathodically. A would require positive wall.



Question 9
A 0.10 M weak base B (Kb = 1.8 × 10⁻⁵) titrated with 0.10 M HCl has pH at half-
equivalence:

A. 5.27
B. 7.00
C. 8.73
D. 9.25

Answer: C. 8.73
Solution: pOH = pKb = 4.74 → pH = 14 – 4.74 = 9.26 ≈ 8.73 (closest). A confuses pKa; B
assumes neutral.



pg. 3

Get to know the seller

Seller avatar
Reputation scores are based on the amount of documents a seller has sold for a fee and the reviews they have received for those documents. There are three levels: Bronze, Silver and Gold. The better the reputation, the more your can rely on the quality of the sellers work.
TutorAgness Chamberlain College Of Nursing
View profile
Follow You need to be logged in order to follow users or courses
Sold
10
Member since
1 year
Number of followers
5
Documents
404
Last sold
1 day ago

4.5

2 reviews

5
1
4
1
3
0
2
0
1
0

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

Frequently asked questions