Analytical Chem istry - DePauw University

420 Analytical Chemistry 2.0141210pH86Figure 9.6 Titration curve for the titration of 50.0 mL of 0.100 MHCl with 0.200 M NaOH. The red points correspond to the datain Table 9.2. The blue line shows the complete titration curve.4200 10 20 30 40 50Volume of NaOH (mL)Ti t r a t i n g a We a k Ac i d w it h a St r o n g Ba s eStep 1: Calculate the volume of titrantneeded to reach the equivalence point.For this example, let’s consider the titration of 50.0 mL of 0.100 M aceticacid, CH 3 COOH, with 0.200 M NaOH. Again, we start by calculatingthe volume of NaOH needed to reach the equivalence point; thusmolesCHCOOH3=molesNaOHVeqM × V = M × Va a b b= MVV = a aM)(50.0 mL)bM= ( 0.100. M= 25.0mL0 200bStep 2: Before adding the titrant, the pHis determined by the titrand, which in thiscase is a weak acid.Before adding NaOH the pH is that for a solution of 0.100 M aceticacid. Because acetic acid is a weak acid, we calculate the pH using themethod outlined in Chapter 6.+ −CH COOH( aq) + HO() l HO ( aq) + CH COO ( aq )3 2 3 3Ka+ −[ HO ][ CH COO ]3 3( x)( x)= =[ CH COOH]0.100− x= 1.75× 10 −53Because the equilibrium constant for reaction9.2 is quite largeK = K a /K w = 1.75 10 9we can treat the reaction as if it goes tocompletion.x = [ HO+ ] = 132 . × 10 −3M3At the beginning of the titration the pH is 2.88.Adding NaOH converts a portion of the acetic acid to its conjugatebase, CH 3 COO – .−−CH COOH( aq) + OH ( aq) → HO() l + CH COO ( aq )3 2 3Any solution containing comparable amounts of a weak acid, HA, and itsconjugate weak base, A – , is a buffer. As we learned in Chapter 6, we cancalculate the pH of a buffer using the Henderson–Hasselbalch equation.9.2