Ch.4 - Chemical Quantities & Aqueous ReactionsWorksheetSee 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: A solution was prepared by mixing 50.00 mL of 0.100 M HNO 3 and 100.00 mL of 0.200 M HNO3. Calculate the molarity of the final solution of nitric acid.

Solution: A solution was prepared by mixing 50.00 mL of 0.100 M HNO 3 and 100.00 mL of 0.200 M HNO3. Calculate the molarity of the final solution of nitric acid.

Problem

A solution was prepared by mixing 50.00 mL of 0.100 M HNO 3 and 100.00 mL of 0.200 M HNO3. Calculate the molarity of the final solution of nitric acid.

Solution

When we mix two solutions of differing concentrations, the final concentration become smaller than the initial concentration because of dilution resulting from the increase in total volume.

We first need to calculate the moles of HNO3 in the original solutions. Recall that molarity is:

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