1. Concentration of Solutions
Chapter 4

2. volume of solution in liters
Molarity
Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
Molarity is one way to measure the concentration of a solution.
moles of solute
volume of solution in liters
Molarity (M) = 2

3. Mixing a Solution
Procedure for preparing L of 1.00 M solution of CuSO4
Weight out 39.9 g of CuSO4
= mol
Add 250 mL of water to volumetric flask
Molarity = .250 mol/.250 L = 1.00 M 3

4. Expressing the concentration of an Electrolyte
When an ionic compound dissociates, the relative concentrations of the ions introduced into the solution depends on the chemical formula of the compound
Example:
1.0 M NaCl
1.0 M Na+
1.0 M Cl-
1 M Na2SO4
2.0 M Na+
1.0 M SO42-

5. Using Molarities in Stoichiometric Calculations5

6. Moles = 2.0 L soln x 0.200 mol HNO3 / 1 L solution
Practice Question
How many moles of HNO3 are in 2.0 L of M HNO3 solution?
Moles = 2.0 L soln x mol HNO3 / 1 L solution
=0.40 mol HNO3

7. Dilution 7

8. Dilution
Solutions of lower concentration can be obtained by adding water
Moles solute in conc soln = moles solute in dilution sln
Therefore, Mconc x Vconc = Mdil x Vdil

9. Practice problem
How many liters of 3.0 M H2SO4 are needed to make .450 L of 0.10 M H2SO4?
Mconc x Vconc = Mdil x Vdil
Moles H2SO4 in dilute solution = L x 0.10 mol/L
= mol = moles H2SO4 in conc solution
L conc soln = mol x 1 L/ 3.0 mol
= .015 L

10. Titration
The analytical technique in which one can calculate the concentration of a solute in a solution.
One reagant has known concentration 10

11. Titration
Indicators are used to determine the equivalence point of the reaction
Point where the neutralization reaction between 2 reactants are complete
Reactants in stoichiometric equivalents are brought together 11

12. Practice Problem
If 45.7 mL of M H2SO4 is required to neutralize a 20.0 mL sample of NaOH solution. What is the concentration of the NaOH solution?
Moles H2SO4 = L x mol H2SO4 /L
= 2.28 x 10-2 mol H2SO4
Balanced Equation: H2SO4 + 2 NaOH  2 H2O + Na2SO4
Moles NaOH = 2.28 x 10-2 mol H2SO4 x 2 mol NaOH/1 mol H2SO4
= 4.56 x 10-2 mol NaOH
Molarity of NaOH = 4.56 x 10-2 mol NaOH/.020 L soln
= 2.28 M NaOH