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
Concentration = moles/volume
= 0.480 M
2.355 g
98.1 g/mol
1 L50 mL x
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
Given volume of A with molarity MA
Determine moles A
Determine moles B
Find target volume of B with molarity MB
Volume Bmol = MV Mole:mole ratio V = mol/MVolume A Moles A Moles 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?
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