Solution Stoichiometry (Review) Titration: HA + BOH  H2O + AB molar mass A molarity HA (M) g A mol HA L of HA g A 1 mol A mol HA 1 L mol-to-mol ratio g B 1 mol B mol BOH 1 L molar mass B molarity BOH (M) g B mol BOH L of BOH 1

Example #1: L of B  L of A HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) Hydrobromic acid is titrated to equivalence with sodium hydroxide. What volume of 2.00 M HBr is needed to react with 10. L of 2.00 M of NaOH ? HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) V = ? L V = 10. L M = 2.00 mol/L M = 2.00 mol/L molarity HA 2.00 mol NaOH 1 mol HBr 1 L HBr x x 10 L NaOH x 1 L NaOH 1 mol NaOH 2.00 mol HBr L of BOH molarity BOH = ____L HBr 10. L HBr

Example #2: L of B  L of A HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) Hydrobromic acid is titrated to equivalence with sodium hydroxide. What volume of 1.50 M HBr is needed to react with 120. L of 1.25 M of NaOH ? HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) V = ? L V = 120. L M = 1.50 mol/L M = 1.25 mol/L molarity HA 1.25 mol NaOH 1 mol HBr 1 L HBr x x 120 L NaOH x 1 L NaOH 1 mol NaOH 1.50 mol HBr L of BOH molarity BOH = ____L HBr 100 L HBr

Example #3: KOH(aq) + HNO3(aq)  H2O(l) + NaNO3(aq) A 20.4 mL solution of KOH was completely neutralized by 48.5 mL of 0.150 M HNO3 . What is the concentration of the KOH? KOH(aq) + HNO3(aq)  H2O(l) + NaNO3(aq) V = 20.4 mL V = 48.5 mL M = ? mol/L M = 0.150 mol/L 0.150 mol HNO3 1 mol KOH 0.00728mol KOH x 0.0485 L HNO3 x = 1 L HNO3 1 mol HNO3 L of HA molarity HA molarity BOH 0.00728 mol KOH = _____ M KOH 0.357 M KOH 0.0204 L KOH

Example #4: L of A  M of B A 159 mL sample of KOH is titrated to equivalence with 100. mL of 1.50 M H2SO4 . Calculate the molarity of the of the KOH solution. H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + K2SO4(aq) V = 0.100 L V = 0.159 L M = 1.50 mol/L M = ? mol/L 1.50 mol H2SO4 2 mol KOH 0.300 mol KOH x = 0.100 L H2SO4 x 1 L H2SO4 1 mol H2SO4 L of HA molarity HA molarity BOH 0.300 mol KOH = _____ M KOH 1.89 M KOH 0.159 L KOH