Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods421−ApH = pK + log [ ]a[ HA]Before the equivalence point the concentration of unreacted acetic acid isinitialmoles CH COOH − molesNaOH added3[ CH COOH]=3totalvolumeMV − MVa a b b=V + VabStep 3: Before the equivalence point, thepH is determined by a buffer containingthe titrand and its conjugate form.and the concentration of acetate ismolesNaOHadded[ CH COO− b b] = = MV3totalvolume V + VFor example, after adding 10.0 mL of NaOH the concentrations ofCH 3 COOH and CH 3 COO – are(0.100 M)(50.0 mL) −(0.200 M)(10.0 mL)[ CH COOH ] =350.0 mL + 10.0 mL= 0.0500 M[ CH COO ]which gives us a pH of3− =( 0.200 M)(10.0 mL)50.0 mL + 10.0 mLa= 0.03330 0333 MpH = 476 . + log . = 458 .0.0500 MbMAt the equivalence point the moles of acetic acid initially present andthe moles of NaOH added are identical. Because their reaction effectivelyproceeds to completion, the predominate ion in solution is CH 3 COO – ,which is a weak base. To calculate the pH we first determine the concentrationof CH 3 COO –molesNaOHadded[ CH COO− ] =3totalvolume( .= 0200 M)(25.0 mL)= 0.0667 M50.0 mL + 25.0 mLNext, we calculate the pH of the weak base as shown earlier in Chapter 6.−−CH COO ( aq) + H O() l OH ( aq) + CHCOOH( aq )3 2 3Step 4: The pH at the equivalence pointis determined by the titrand’s conjugateform, which in this case is a weak base.Alternatively, we can calculate acetate’s concentrationusing the initial moles of acetic acid;thus−initialmoles CH COOH3[ CH COO ]3 =totalvolume( 0.100 M)(50.0 mL)=50.0 mL + 25.0 mL= 0.0667 M