Concentration of Solutions

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 litersMolarity (M) =

Mixing a Solution

Procedure for preparing 0.250 L of 1.00 M solution of CuSO4

• Weight out 39.9 g of CuSO4

= 0.250 mol

• Add 250 mL of water to volumetric flask

• Molarity = .250 mol/.250 L = 1.00 M

Expressing the concentration of an Electrolyte

• When an ionic compound dissolves, 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-

Using Molarities inStoichiometric Calculations

Practice Question

• How many moles of HNO3 are in 2.0 L of 0.200 M HNO3 solution?

Moles = 2.0 L soln x 0.200 mol HNO3 / 1 L solution

=0.40 mol HNO3

Dilution

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

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 = 0.450 L x 0.10 mol/L

= 0.045 mol = moles H2SO4 in conc solution L conc soln = 0.045 mol x 1 L/ 3.0 mol

= .015 L

Titration• The analytical

technique in which one can calculate the concentration of a solute in a solution.

• One reagant has known concentration

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

Practice ProblemIf 45.7 mL of 0.500 M H2SO4 is required to neutralize a

20.0 mL sample of NaOH solution. What is the concentration of the NaOH solution?

Moles H2SO4 = .0457 L x 0.500 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