Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
All Chapters
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

What is the molarity of a NaOH solution if 39.1 mL of a 0.112 M H2SO4 solution is required to neutralize a 25.0-mL sample of the NaOH solution (as in a titration)? 

A) 54.7 M 

B) 0.224 M 

C) 0.350 M 

D) 0.143 M 

E) 0.0716 M


We’re being asked to calculate the molarity of NaOH (strong base) from the given the molarity and volume of H2SO4

This means that 25.00 mL of NaOH required 39.1 mL of 0.112 M H2SO4 to reach the equivalence point.

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