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?