Chapter 16.4-16.5 Titrations of Weak Acids and Bases Objectives ◦Separate the titration curve for a weak acid into regions in which a single equilibrium dominates ◦Estimate the pH of mixtures of a weak acid and strong base or weak base and strong acid. ◦Calculate the pH in the titration of a weak acid with strong base. ◦Calculate the pH in the titration of a weak base with strong acid. October 13th, 2015 Titration Calculation Strategy Break titration into 4 zones: • pH before any titrant is added. • pH after titrant is added but before equivalence point. • pH at equivalence point. • pH after equivalence point. Titration of Weak Acid with Strong Base Strong base region Buffer region pH Equivalence point Weak acid 0 5 10 15 20 25 Volume of NaOH added, mL 30 Titration of a Weak Base with a Strong Acid 14 Weak base 12 10 8 pH Buffer region 6 4 Equivalence point 2 Strong acid region 0 0 5 10 15 20 Volume of HCl added 25 30 In Class Example Calculate the pH in the titration of 20.00 mL of 0.500 M formic acid (Ka = 1.8 x 10-4) with 0.500 M NaOH after 0, 10.00, 20.00, and 30.00 mL of base have been added. Graph the titration curve. The reaction is: HCOOH + OH- HCOO- + H2O Zone 1: pH before any titrant is added 0 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH Ka = 1.8 x 10-4 No base has been added, so the solution only contains a weak acid. Therefore, use our knowledge of how to calculated pH from Ka values. H3 O+ [HCOO− ] Ka = [HCOOH] Zone 2: pH after titrant is added but before equivalence point The solution is a mixture of the weak acid and its conjugate base – a buffer. The pH can be determined using a sRfc table and the Henderson-Hasselbalch equation. nb pH = pKa + log na sRfc Table for Zone 2 10.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH HCOOH + OH- HCOO- + H2O s, mmol R, mmol f, mmol c, M Zone 3: pH at equivalence point Use the sRfc table to find the equivalence point. The equivalence point occurs when the mmols of titrant added equals the initial mmols of HCOOH. At the equivalence point, only weak base is present in solution, so the pH can be calculated from our knowledge of Kb HCOOH [OH− ] Kb = [HCOO− ] sCrf Table at equivalence point 20.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH HCOOH + OH- HCOO- + H2O s, mmol R, mmol f, mmol c, M Zone 4: pH after equivalence point After the equivalence point there is excess strong base. There is also weak base around, but in Chapter 15.7 we learned that the stronger base dominates the pH and the weak base can be ignored. Use the sRfc table to determine the amount of strong base and use our knowledge of pKw to find pH pKw = 14 = pH + pOH sCrf Table at equivalence point 30.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH HCOOH + OH- HCOO- + H2O s, mmol R, mmol f, mmol c, M Student Example Calculate the pH in the titration of 12.00 mL of 0.100 M CH3COO- with 0.200 M HCl after 0, 3.00, 6.00 and 9.00 mL of base have been added. Kb = 5.510-10
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