Question

Consider the titration of 35.0 mL of 0.143-M of KX with 0.165-M HCl. The pKa of HX = 6.83. Give all pH values to 0.01 pH units.

a) What is the pH of the original solution before addition of any acid? pH =

b) How many mL of acid are required to reach the equivalence point? VA = mL

c) What is the pH at the equivalence point? pH =

d) What is the pH of the solution after the addition of 18.8 mL of acid? pH =

e) What is the pH of the solution after the addition of 36.4 mL of acid? pH =

Answer 1

Assume

KX --> K+ + X-

assume X- is B so X-+H = BH+

B + H2O <-> HB+ + OH-

Kb = [HB+][OH-]/[B]

pKb = 14-pKa = 14-6.83 = 7.17

Kb = 10^-7.17

so....

a)

no volume of acid

B + H2O <-> HB+ + OH-

Kb = [HB+][OH-]/[B]

in equilibrium

for 1 mol of OH- we hava always 1 mol of HB+

[HB+] = [OH-] = x (the dissociation fraction of acid)

[B] = M-x = 0.07-x (Account for dissociation!)

10^-7.17 = x*x/(0.143-x)

solve with quadratic formula

x = [OH-] = 9.82*10^-5

pOH = -log(Oh-) = -log( 9.82*10^-5) = 4.01

pH = 14-pOH = 14-4.01= 9.99 = 10

b)

Required volume for equivalence

mmol of KX = MV = (0.143)(35) = 5.005

mmol of HCl required = 5.005

V HCl = mmol/M = 5.005 / (0.165) = 30.33 mL

c)

Vtota in equiv. = 30.33+35 = 65.33 mL

[X-] = 5.005/65.33 = 0.0766

HX + H2O <-> H3O + X- is formed

Ka = [H+][X-]/[HX]

(10^-6.83) = x*x/(0.0766-x)

x = [H+] = 1.786*10^-4

pH = -log(1.786*10^-4) = 3.75

d)

mmol of KX = MV = 0.143*35 = 5.005

mmol of HCl = MV = 18.8*0.165 = 3.102

after reaction

mmol of X- left = 5.005-3.102 = 1.903

mmol of HX formed = 3.102

this is a buffer

pH = pKa + log(X-/HX)

pH = 6.83 + log(1.903/3.102)

ph = 6.617

e)

mmol of HCl = MV = (36.4*0.165) = 6.006

mmol of KX = 5.005

mmol of H+ left = 6.006-5.005 = 1.001

V total = 36.4+35 = 71.4 mL

[H+] = mmol/V = 1.001/71.4 = 0.0140

pH = -log(0.0140) = 1.85