Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.

Determine the acid dissociation constant for this weak acid at 25°C.

A. 5.81x10-4

B. 1.85x10-4

C. 1.05x10-9

D. 1.58x10-4

E. 5.01x10-9


We’re being asked to determine the acid dissociation constant (Ka) for a weak monoprotic acid

Recall that at the equivalence point of a titration:

moles acid=moles base

Also, recall that moles = molarity × volume

This means:


The weak acid is monoprotic so it ha only one H+ to donate and one equivalence point:

HA H+ + A-        Ka                  Equivalence Point

            (MA)acid = (MV)base

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