Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry283[ I ] = K = 83 . × 10 = 91 . × 10− −17 −9spcorresponding to a pI of 8.04. The following session shows the functionin action.> pI =c(4, 5, 6, 7, 8)> eval(pI)pI error1 4 -2.562356152 5 -0.166209303 6 0.073371014 7 0.097348245 8 0.09989073> pI =c(5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 6.0)> eval(pI)pI error1 5.1 -0.111446582 5.2 -0.067941053 5.3 -0.033364754 5.4 -0.005681165 5.5 0.015715496 5.6 0.033089297 5.7 0.046859378 5.8 0.057792149 5.9 0.0664747510 6.0 0.07337101> pI =c(5.40, 5.41, 5.42, 5.43, 5.44, 5.45, 5.46, 5.47, 5.48, 5.49, 5.50)> eval(pI)pI error1 5.40 -0.00568116052 5.41 -0.00307154843 5.42 0.00023103694 5.43 -0.00051348985 5.44 0.00282818786 5.45 0.00523709807 5.46 0.00747581818 5.47 0.00962603709 5.48 0.011710549810 5.49 0.013738729111 5.50 0.0157154889The error function is closest to zero at a pI of 5.42. The concentration ofI – at equilibrium, and the molar solubility of AgI, is 3.8 × 10 –6 mol/L,which agrees with our earlier solution to this problem.Click here to return to the chapterM