1. Unit 10: Solution stoichiometry

2. Learning objectives Calculate molarity and dilution factors
Use molarity in solution stoichiometry problems
Apply solution stoichiometry to acid-base titrations

3. 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

4. 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

5. Example What is molarity of 50 ml solution containing 2.355 g H2SO4?
Molar mass H2SO4 = 98.1 g/mol
Moles H2SO4 = mol
Volume of solution = L
Concentration = moles/volume
= M

6. What is concentration of solution containing 60 g NaOH in 1.5 L

7. Moles before = moles after:
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

8. What is concentration if 2 L of 6 M HCl is diluted to 12 L?

9. How much water must be added to make a 2 M solution from 100 mL of 6M solution?

10. 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

11. 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

12. Example
How much M NaHCO3 is required to neutralize 18.0 mL of M HCl?