Acid-Base Mixing and Titrations

Question

Consider the titration of 45.0 mL of 0.117-M of KX with 0.147-M HCl. The pK_a of HX = 10.07. Give all pH values to 0.01 pH units.

I have the pH of original solution before adding acid at 11.56 and ml of acid to reach equivalence point at 36.05.

I don't understand these parts of the question.

c) What is the pH at the equivalence point?

d) What is the pH of the solution after the addition of 25.1 mL of acid?

e) What is the pH of the solution after the addition of 43.0 mL of acid?

Thank you

Answer 1

X^- + H2O ==> HX + OH^-

X^- is acting as a base (conjugate base of HX) so Kb = 14 - pKa = 3.93 and Kb = 1.17x10^-4

Kb = 1.17x10^-4 = [HX][OH-]/[X-] = (x)(x) / 0.117 - x and ignoring x in denominator we have..

x² = 1.37x10^-5

x = 1.17x10^-3 M = [OH-]

pOH = 2.93

pH = 14 - pOH

pH = 11.07 (I don't get what you get)

ml of acid to reach equivalence?

45.0 ml X- x 1 /1000 ml x 0.117 mol/L = 0.005265 moles X- = 0.005265 moles OH- produced

(x L)(0.147 mol/L) = 0.005265 moles and x = 35.82 mls of HCl (again, I don't get what you get)

pH @ equivalence? At the equivalence point, all the X- has reacted with H+ from HCl leaving only KCl and HX. KCl is neutral so the pH @ equivalence is determined by [HX] formed.

KX + HCl ==> HX + KCl

moles HX formed = 0.005265 (from that many moles of KX and HCl)

Total volume = 45 ml + 35.82 ml = 80.82 ml = 0.08082 L

[HX] = 0.005265 mol/0.08082 L = 0.06514 M

pKa HX = 10.07 and Ka = 8.51x10^-11

Ka = 8.51x10^-11 = [H+][X-]/[HX] = (x)(x)/0.06514

x2 = 5.54x10^-12

x = [H+] = 2.35x10^-6 M

pH @ equivalence = -log 2.35x10-6 = 5.63

To do (d) and (e) calculate final volume and moles HCl added. This will be large compared to the [H+] just calculated, so pH will be essentially that resulting from diluting the HCl.

Hope this makes sense.