Analytical Chem istry - DePauw University

244 Analytical Chemistry 2.0pHOH 2 N CH C O–CH 3pK a1= 9.867O+ H3 N CH C O–CH 3pK a2= 2.348O+H 3 N CH C OHCH 3L –HLH 2 L +Figure 6.13 Ladder diagram for alanine.pH o f 0.10 M Al a n i n e Hy d r o c h l o r i d e (H 2 L + )Alanine hydrochloride is a salt of the diprotic weak acid H 2 L + and Cl – .Because H 2 L + has two acid dissociation reactions, a complete systematicsolution to this problem is more complicated than that for a monoproticweak acid. The ladder diagram in Figure 6.13 helps us simplify the problem.Because the areas of predominance for H 2 L + and L – are so far apart, we canassume that a solution of H 2 L + is not likely to contain significant amountsof L – . As a result, we can treat H 2 L + as though it is a monoprotic weak acid.Calculating the pH of 0.10 M alanine hydrochloride, which is 1.72, is leftto the reader as an exercise.pH o f 0.10 M So d i u m Al a n i n a t e (L – )The alaninate ion is a diprotic weak base. Because L – has two base dissociationreactions, a complete systematic solution to this problem is morecomplicated than that for a monoprotic weak base. Once again, the ladderdiagram in Figure 6.13 helps us simplify the problem. Because the areasof predominance for H 2 L + and L – are so far apart, we can assume that asolution of L – is not likely to contain significant amounts of H 2 L + . As aresult, we can treat L – as though it is a monoprotic weak base. Calculatingthe pH of 0.10 M sodium alaninate, which is 11.42, is left to the reader asan exercise.pH o f 0.1 M Al a n i n e (HL)Finding the pH of a solution of alanine is more complicated than our previoustwo examples because we cannot ignore the presence of both H 2 L +and L – . To calculate the solution’s pH we must consider alanine’s acid dissociationreaction+ −HL( aq) + HO() l HO ( aq) + ( aq )2 3Land its base dissociation reaction− +HL( aq) + HO() l OH ( aq) + HL( aq )2 2As always, we must also consider the dissociation of water2H O() l H O + ( aq) + OH− ( aq )2 3This leaves us with five unknowns—[H 2 L + ], [HL], [L – ], [H 3 O + ], and[OH – ]—for which we need five equations. These equations are K a2 andK b2 for alanine+ −[ HO ][ L ]=[ HL]K a23