Solution stoichiometry
Volumetric calculations
Acid-base titrations
Learning objectives



Calculate molarity and dilution factors
Use molarity in solution stoichiometry
problems
Apply solution stoichiometry to acid-base
titrations
Solution stoichiometry

In solids, moles are obtained by dividing mass
by the molar mass

In liquids, it is necessary to convert volume
into moles using molarity
Molarity (M)
Molarity (M) = Moles of solute/Liters of solution

Stoichiometric calculations are facile

Amounts of solution required are volumetric

Concentration varies with T

Amount of solvent requires knowledge of
density
Example

What is molarity of 50 ml solution containing
2.355 g H2SO4?

Molar mass H2SO4 = 98.1 g/mol
Moles H2SO4 = 0.0240 mol

Volume of solution = 0.050 L

2.355 g
98.1 g/mol
50 mL x

Concentration = moles/volume
= 0.480 M
1L
1000 mL
0.0240 mol
0.050 L
What is concentration of solution
containing 60 g NaOH in 1.5 L
Dilution

More dilute solutions are prepared from concentrated
ones by addition of solvent
Moles before = moles after:
M1V1 = M2V2
Molarity of new solution M2 = M1V1/V2
To dilute by factor of ten, increase volume by factor of ten

Do molarity exercises
What is concentration if 2 L of 6 M
HCl is diluted to 12 L?
How much water must be added to make a 2
M solution from 100 mL of 6M solution?
Solution stoichiometry

How much volume of one solution to react with
another solution




Volume A
Given volume of A with molarity MA
Determine moles A
Determine moles B
Find target volume of B with molarity MB
mol = MV
Moles A
Mole:mole ratio
Moles B
V = mol/M
Volume B
Titration



Use a solution of known concentration to determine
concentration of an unknown
Must be able to identify endpoint of titration to know
stoichiometry
Most common applications with acids and bases
Example

How much 0.125 M NaHCO3 is required to neutralize
18.0 mL of 0.100 M HCl?