Analytical Chem istry - DePauw University

504 Analytical Chemistry 2.0Step 3: Calculate pCl at the equivalencepoint using the K sp for AgCl to calculatethe concentration of Cl – .Table 9.18 Titration of 50.0 mL of 0.0500 M NaCl with 0.100 M AgNO 3Volume of AgNO 3 (mL) pCl Volume of AgNO 3 (mL) pCl0.00 1.30 30.0 7.545.00 1.44 35.0 7.8210.0 1.60 40.0 7.9715.0 1.81 45.0 8.0720.0 2.15 50.0 8.1425.0 4.89which corresponds to a pCl of 1.60.At the titration’s equivalence point, we know that the concentrationsof Ag + and Cl – are equal. To calculate the concentration of Cl – we use theK sp expression for AgCl; thus+ − −K = [ Ag ][ Cl ] = ( x)( x) = 18 . × 10 10spStep 4: Calculate pCl after the equivalencepoint by first calculating the concentrationof excess AgNO 3 and then calculatingthe concentration of Cl – using the K spfor AgCl.Solving for x gives [Cl – ] as 1.3 × 10 –5 M, or a pCl of 4.89.After the equivalence point, the titrant is in excess. We first calculate theconcentration of excess Ag + and then use the K sp expression to calculate theconcentration of Cl – . For example, after adding 35.0 mL of titrant+ molesAg added−initial molesCl[ Ag ] =total volume+ −M V=V( 0. 100 M)(35. 0 mL) −( 0. 0500 M)( 50. 0 mL)=50. 0 mL + 350 . mLAg Ag Cl ClCl− M V+ VAg= 118 . × 10 −2MK−− sp 18 . × 10[ Cl ] = =+[ Ag ] 118 . × 1010−2= 15 . × 10−8Mor a pCl of 7.81. Additional results for the titration curve are shown inTable 9.18 and Figure 9.43.1086pCl4Figure 9.43 Titration curve for the titration of 50.0mL of 0.0500 M NaCl with 0.100 M AgNO 3 . The redpoints corresponds to the data in Table 9.18. The blueline shows the complete titration curve.200 10 20 30 40 50Volume of AgNO 3