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: 38. 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: 38. 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

38. 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

We’re being asked to calculate the molarity (M) of nitric acid (HNO3) when 50.00 mL of 0.100 M HNO3 and 100.00 mL of 0.200 M HNO3 are mixed.


Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters). In other words:



We first need to determine the moles of HNO3 present in each initial solution. This is because we mixed two solutions containing HNO3 and the moles of HNO3 in the new solution is simply the sum of the moles of HNO3 in the original solutions.


Note that we need to convert mL to L since molarity is in mol/L.

View the complete written solution...