Analytical Chem istry - DePauw University

436 Analytical Chemistry 2.0and the pOH is 5.3. The titrand’s pH ispH = pK − pOH = 14 − 53 . = 87 .wand the change in the titrand’s pH as the titration goes from 90% to 110%of V eq is∆pH = 87 . − 53 . = 34 .If we carry out the same titration in a nonaqueous solvent with a K s of1.010 –20 , the pH after adding 45.0 mL of NaOH is still 5.3. However,the pH after adding 55.0 mL of NaOH isIn this case the change in pHpH = pK − pOH = 20 − 53 . = 14.7s∆pH = 14. 7− 5. 3=9.4is significantly greater than that obtained when the titration is carried outin water. Figure 9.17 shows the titration curves in both the aqueous andthe nonaqueous solvents.Another parameter affecting the feasibility of an acid–base titration isthe titrand’s dissociation constant. Here, too, the solvent plays an importantrole. The strength of an acid or a base is a relative measure of the easetransferring a proton from the acid to the solvent, or from the solvent tothe base. For example, HF, with a K a of 6.8 10 –4 , is a better proton donorthan CH 3 COOH, for which K a is 1.75 10 –5 .The strongest acid that can exist in water is the hydronium ion, H 3 O + .HCl and HNO 3 are strong acids because they are better proton donors thanH 3 O + and essentially donate all their protons to H 2 O, leveling their acid2015(b)pH10(a)5Figure 9.17 Titration curves for 50.0 mL of 1.0 10 –4M HCl using 1.0 10 –4 M NaOH in (a) water,K w = 1.0 10 –14 , and (b) a nonaqueous solvent,K s = 1.0 10 – 20 .00 20 40 60 80 100Volume of NaOH(mL)