12.07.2015 Views

Analytical Chem istry - DePauw University

Analytical Chem istry - DePauw University

Analytical Chem istry - DePauw University

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Chapter 9 Titrimetric Methods503Calcium nitrate, Ca(NO 3 ) 2 , was used as the titrant, forming a precipitateof CaCO 3 and CaSO 4 . The titration’s end point was signaled by notingwhen the addition of titrant ceased to generate additional precipitate. Theimportance of precipitation titrimetry as an analytical method reached itszenith in the nineteenth century when several methods were developed fordetermining Ag + and halide ions.9E.1 Titration CurvesA precipitation titration curve follows the change in either the titrand’s orthe titrant’s concentration as a function of the titrant’s volume. As we havedone with other titrations, we first show how to calculate the titration curveand then demonstrate how we can quickly sketch a reasonable approximationof the titration curve.Ca l c u l at i n g t h e Ti t r a t i o n Cu r v eLet’s calculate the titration curve for the titration of 50.0 mL of 0.0500 MNaCl with 0.100 M AgNO 3 . The reaction in this case is+ −Ag ( aq) + Cl ( aq) AgCl()sBecause the reaction’s equilibrium constant is so largeK− − −= ( K ) = (. 18× 10 ) = 56 . × 10sp1 10 1 9we may assume that Ag + and Cl – react completely.By now you are familiar with our approach to calculating a titrationcurve. The first task is to calculate the volume of Ag + needed to reach theequivalence point. The stoichiometry of the reaction requires thatStep 1: Calculate the volume of AgNO 3needed to reach the equivalence point.molesAgSolving for the volume of Ag +Veq= moles Cl+ −M × V = M × VAg Ag Cl ClM VCl ClM)(50.0 mL)= V = = ( 0.0500 AgM (0.100 M)Ag= 25.0mLshows that we need 25.0 mL of Ag + to reach the equivalence point.Before the equivalence point the titrand, Cl – , is in excess. The concentrationof unreacted Cl – after adding 10.0 mL of Ag + , for example, is− +− initialmoles Cl − molesAg added M V[ Cl ] ==total volumeV( 0. 0500 M)(50. 0mL) −( 0. 100 M)( 10. 0 mL)=50. 0 mL + 100 . mLCl Cl Ag AgCl− M V+ VAg= 250 . × 10 −2MStep 2: Calculate pCl before the equivalencepoint by determining the concentrationof unreacted NaCl.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!