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

Solution: Consider the given graph. If the given graph shows the titration of 23.00 mL of an HNO3 solution of an unknown concentration with a 0.135 M LiOH solution, what is the molarity of the acid solution?

Solution: Consider the given graph. If the given graph shows the titration of 23.00 mL of an HNO3 solution of an unknown concentration with a 0.135 M LiOH solution, what is the molarity of the acid solution?

Problem

Consider the given graph. If the given graph shows the titration of 23.00 mL of an HNO3 solution of an unknown concentration with a 0.135 M LiOH solution, what is the molarity of the acid solution?

Solution

We’re being asked to calculate the molarity of HNO3 (strong acid) from the given titration data with LiOH (strong base)


Recall that in the titration of a strong acid and strong base, the equivalence point occurs at pH 7. From the titration curve, we can determine the volume of LiOH:



This means that 23.00 mL of HNO3 required 20.00 mL of 0.135 M LiOH to reach the equivalence point.


View the complete written solution...