Ch109_LectureNotes

balance as in acid, then:
1. Add enough OH^- (equal numbers to both sides) to cancel the H⁺. (This is necessary because there will not be protons present in a basic solution!). (There are a couple of other conventions for balancing in basic solutions. If you are familiar with another and prefer it, you may use it instead.)
2. Combine the H⁺ and OH^- on the appropriate side of the equation to give waters.
3. Go back and cancel waters which appear on both sides to give the final equation.

Molarity and the Composition of Solutions

The most commonly used concentration term in chemistry is molarity = 1 mole of solute dissolved in enough solvent to give 1 L = M = mol/L. This is the most popular concentration unit at least in part because of the convenience of making molar solutions with volumetric flasks (show flask).

Note that molarity has various meanings in describing solutions. Commonly molarity refers to the amount of substance dissolved in one liter of liquid to give a solution. Thus if 158.52 g (1 mole) of strontium chloride is dissolved in enough water to give one liter we will have a 1 L solution of strontium chloride. Note however, that while the solution is 1M in Sr²⁺, it is 2 M in Cl^- and 3 M in terms of particles!

Making molar solutions

Determining Molar Concentration

Dilution problems

Acid-Base Reactions

Neutralization

When we combine equal numbers of moles of hydrogen ion and hydroxide ion a neutralization occurs. That is, there is no reactive component left, all of the acid has been consumed by all of the base, and water has been synthesized.

First find out the moles of each:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles;
base = (25.0 mL)(1 L/1000 mL)(0.50 mole/L) = 1.25 x 10^-2moles.

Since the amounts are identical they will completely neutralize each other. That is they will react completely and both acid and base will be consumed with none left over.
Since the ratio in the equation is 1:1:1 acid:base:water, 1.25 x 10^-2moles of water are produced.

Consider the reaction of a weak acid and strong base: 50.0 mL of 0.25 M acetic acid is reacted with 18.0 mL of 0.50 M sodium hydroxide. Find the number of moles of each of the reactants and products after reaction.

First recall that for the reaction of a weak acid and a strong base the reaction will go until one of the reactants is completely consumed.

Writing the net ionic reaction we get:

CH₃COOH + OH^- H₂O + CH₃COO^-

First find moles of each reactant:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles

base = (18.0 mL)(1 L/1000 mL)(0.50 mole/L) = 0.900 x 10^-2moles

After reaction all of the base is consumed, so:

base = 0

acetate = 0.900 x 10^-2moles

acid = 1.25 x 10^-2moles - 0.900 x 10^-2moles = 0.35 x 10^-2moles

Water synthesized = 0.900 x 10^-2moles.

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 12: 14 February

PREVIOUS

Balancing Redox Equations, cont.

Basic Solution: