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 H 2 SO 4 ? Molar mass H 2 SO 4 = 98.1 g/mol Molar mass H 2 SO 4 = 98.1 g/mol Moles H 2 SO 4 = 0.0240 mol Moles H 2 SO 4 = 0.0240 mol Volume of solution = 0.050 L Volume of solution = 0.050 L Concentration = moles/volume Concentration = moles/volume = 0.480 M = 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: M 1 V 1 = M 2 V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 Molarity of new solution M 2 = M 1 V 1 /V 2 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 M A Given volume of A with molarity M A Determine moles A Determine moles A Determine moles B Determine moles B Find target volume of B with molarity M B Find target volume of B with molarity M B

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 NaHCO 3 is required to neutralize 18.0 mL of 0.100 M HCl?