Question

Calculate the pH during the titration of 40.00 mL of 0.100 M propionic acid (Ka = 1.3 x 10-5) with 0.100 M NaOH at the following volumes:
A) 0.00 mL of base added

B) 7.50 mL of base added

C) 20.00 mL of base added

D) 40.00 mL of based added

E) 45 mL of base added

Answer 1

A)when 0.0 mL of NaOH is added

HA dissociates as:

HA -----> H+ + A-

0.1 0 0

0.1-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.3*10^-5)*0.1) = 1.14*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.3*10^-5 = x^2/(0.1-x)

1.3*10^-6 - 1.3*10^-5 *x = x^2

x^2 + 1.3*10^-5 *x-1.3*10^-6 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.3*10^-5

c = -1.3*10^-6

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 5.2*10^-6

roots are :

x = 1.134*10^-3 and x = -1.147*10^-3

since x can't be negative, the possible value of x is

x = 1.134*10^-3

use:

pH = -log [H+]

= -log (1.134*10^-3)

= 2.9455

Answer: 2.95

B)when 7.5 mL of NaOH is added

Given:

M(HA) = 0.1 M

V(HA) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 7.5 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 7.5 mL = 0.75 mmol

We have:

mol(HA) = 4 mmol

mol(NaOH) = 0.75 mmol

0.75 mmol of both will react

excess HA remaining = 3.25 mmol

Volume of Solution = 40 + 7.5 = 47.5 mL

[HA] = 3.25 mmol/47.5 mL = 0.0684M

[A-] = 0.75/47.5 = 0.0158M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.886+ log {1.579*10^-2/6.842*10^-2}

= 4.249

Answer: 4.25

C)when 20.0 mL of NaOH is added

Given:

M(HA) = 0.1 M

V(HA) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 20 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HA) = 4 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

excess HA remaining = 2 mmol

Volume of Solution = 40 + 20 = 60 mL

[HA] = 2 mmol/60 mL = 0.0333M

[A-] = 2/60 = 0.0333M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.886+ log {3.333*10^-2/3.333*10^-2}

= 4.886

Answer: 4.89

D)when 40.0 mL of NaOH is added

Given:

M(HA) = 0.1 M

V(HA) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 40 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 40 mL = 4 mmol

We have:

mol(HA) = 4 mmol

mol(NaOH) = 4 mmol

4 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 4 mmol

Volume of Solution = 40 + 40 = 80 mL

Kb of A- = Kw/Ka = 1*10^-14/1.3*10^-5 = 7.692*10^-10

concentration ofA-,c = 4 mmol/80 mL = 0.05M

A- dissociates as

A- + H2O -----> HA + OH-

0.05 0 0

0.05-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((7.692*10^-10)*5*10^-2) = 6.202*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.202*10^-6 M

[OH-] = x = 6.202*10^-6 M

use:

pOH = -log [OH-]

= -log (6.202*10^-6)

= 5.2075

use:

PH = 14 - pOH

= 14 - 5.2075

= 8.7925

Answer: 8.79

E)when 45.0 mL of NaOH is added

Given:

M(HA) = 0.1 M

V(HA) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 45 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 45 mL = 4.5 mmol

We have:

mol(HA) = 4 mmol

mol(NaOH) = 4.5 mmol

4 mmol of both will react

excess NaOH remaining = 0.5 mmol

Volume of Solution = 40 + 45 = 85 mL

[OH-] = 0.5 mmol/85 mL = 0.0059 M

use:

pOH = -log [OH-]

= -log (5.882*10^-3)

= 2.2304

use:

PH = 14 - pOH

= 14 - 2.2304

= 11.7696

Answer: 11.77