Titration Problem!!?

Question

I really don't get this. I have tried to do what I could on my own with no luck! Help please! It looks lengthy so any help is appreciated! Show steps so I can get a understanding!

1. Consider the titration of 35.0 mL of 0.165-M of KX with 0.078-M HCl. The pKa of HX = 5.21. Give all pH values to 0.01 pH units.
a) What is the pH of the original solution before addition of any acid?
b) How many mL of acid are required to reach the equivalence point?
c) What is the pH at the equivalence point?
d) What is the pH of the solution after the addition of 37.0 mL of acid?
e) What is the pH of the solution after the addition of 88.8 mL of acid?

Answer 1

I have helped you previously on a few almost identical problems:
http://answers.yahoo.com/question/index;_ylt=Agm.LoWQ6WNAHlYvGhjhMzbsy6IX;_ylv=3?qid=20071106193516AAJfWjO&show=7#profile-info-JnLbNYviaa
Here I may be able to say more clearly step-by-step.
Of course there is no such true element "X". Hence "KX" is used here to represent a salt, such that "X-" is the conjugated base of acid "HX", and "KX" in solution would be completely dissociate into K+ and X-.
First write the reaction equation:
HX <==> H+ + X-, pKa = 5.21 and Ka = 10^-5.21....(1)

a) Ka = [H+]*[X-] / [HX] May be used to find [H+], but we do not know [HX].
Re-write the reaction as:
X- + H2O <==> HX + OH-
Thus we know [HX] = [OH-]. This property can be used together to solve the problem.
Ka = [H+]*[X-] / [HX]
Ka/Kw = [H+]*[X-] / [HX]*[H+]*[OH-]
= [X-] / [HX]*[OH-]
= [X-] / [OH-]^2
Thus [OH-]^2 = (Kw/Ka)*[X-] = {(10^-14)/(10^-5.21)}*0.165 = 2.676x10^-10
[OH-] = 1.636x10^-5
pH = 14 + log([OH-]) = 14+log(1.636x10^-5) = 9.21

b) (35.0mL)*(0.165M)/(0.078M) = 74 mL

c) The reaction is:
KX + HCl <==> KCl + HX
At the equivalence point, all KX and HCl just "completely reacted" thus only HX and KCl left, and KCl is nutral and spectator thus to be ignored.
HX concentration post titration is:
(35.0mL)*(0.165M) / (35.0mL + 74 mL) = 0.0530M
From (1), we have:
Ka = 10^-5.21 = [H+]*[X-] / [HX] = [H+]^2 / 0.0530
[H+] = sqrt(0.0530*10^-5.21) = 0.000572 (M)....(2)
pH = -log(0.000572) = 3.24

d) The contribution to [H+] from this additional 37.0 mL of acid is:
37.0*0.078/(37.0+35.0+74) = 0.0198 (M)
which is much greater than (2). In fact, with this additional [H+], the self dissociation of HX as shown in (1) would be much smaller than (2). Hence pH is determined by this additional acid and:
pH = -log(0.0198) = 1.70

e) 88.8*0.078/(88.8+35.0+74) = 0.0350 (M)
pH = -log(0.0350) = 1.46