Section 3: Titrations Titration Calculations Print 1, 3-5, 7-8

ACID + BASE  SALT + WATER Neutralization Reactions (REVIEW) Neutralization – rxn of acid and base making a salt and water ACID + BASE  SALT + WATER = HCl + NaOH  NaCl + HOH H2O

Titration MAVA = MBVB moles A = moles B (video clip) The analytical technique used to calculate the moles of an unknown. MAVA = MBVB buret titrant moles A = moles B known vol. (V) known conc. (M) 1:1 ratio ONLY!!! Safari Montage (2 min) Titration analyte known vol. (V) unknown conc. (M)

equivalence point: end point: equal stoichiometric amounts (moles) that react completely mol H+ = mol OH– end point: indicator permanently changes color

Pink in BASE Phenolphthalein Clear in ACID

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 6

Example #1a: 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

MAVA = MBVB Example #1b: L of B  L of A HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) V = ? L V = 120. L M = 1.50 mol/L M = 1.25 mol/L 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 MAVA = MBVB 1:1 ratio ONLY!!! (1.50 M) x VA = (1.25 M) x (120. L) VA = (1.25 M) x (120. L) = (1.50 M) 100 L HBr

MAVA = MBVB Example #2: 1:1 ratio ONLY!!! 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? (0.150 M) x (48.5 mL) = MB x (20.4 mL) (0.150 M) x (48.5 mL) = MB = (20.4 mL) 0.357 M KOH

Example #3: 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

Quick Quiz! 1) When the moles of acid equals the moles bases you have reached the… A) titration point B) end point C) equivalence point D) neutralization point 11

Quick Quiz. 2) If 20. mL of 0.50 M HCl is titrated to equivalence by 40. mL of NaOH, then the NaOH concentration must be ___. A) 2.0 M B) 1.0 M C) 0.50 M D) 0.25 M moles of H+ = moles of OH– 0.020 L 0.50 M 0.040 L ??? M 12