1. Solution stoichiometry Volumetric calculations Acid-base titrations

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) 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 = 0.0240 mol • Volume of solution = 0.050 L • Concentration = moles/volume = 0.480 M

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

7. 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 0.125 M NaHCO3 is required to neutralize 18.0 mL of 0.100 M HCl?