Question

A weak acid, acetic acid (0.100M, 50.0mL, Ka=1.8x 10°) was titrated with 0.100M NaOH. Calculate the pH after 0, 10, 25, 50 and 60mL of NaOH was added.

Answer 1

1)when 0.0 mL of NaOH is added

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.1 0 0

0.1-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*0.1) = 1.342*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)

1.8*10^-5 = x^2/(0.1-x)

1.8*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

use:

pH = -log [H+]

= -log (1.333*10^-3)

= 2.8753

Answer: 2.88

2)when 10.0 mL of NaOH is added

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 10 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 10 mL = 1 mmol

We have:

mol(CH3COOH) = 5 mmol

mol(NaOH) = 1 mmol

1 mmol of both will react

excess CH3COOH remaining = 4 mmol

Volume of Solution = 50 + 10 = 60 mL

[CH3COOH] = 4 mmol/60 mL = 0.0667M

[CH3COO-] = 1/60 = 0.0167M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {1.667*10^-2/6.667*10^-2}

= 4.143

Answer: 4.14

3)when 25.0 mL of NaOH is added

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(CH3COOH) = 5 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess CH3COOH remaining = 2.5 mmol

Volume of Solution = 50 + 25 = 75 mL

[CH3COOH] = 2.5 mmol/75 mL = 0.0333M

[CH3COO-] = 2.5/75 = 0.0333M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {3.333*10^-2/3.333*10^-2}

= 4.745

Answer: 4.75

4)when 50.0 mL of NaOH is added

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 50 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 50 mL = 5 mmol

We have:

mol(CH3COOH) = 5 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 5 mmol

Volume of Solution = 50 + 50 = 100 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 5 mmol/100 mL = 0.05M

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.05 0 0

0.05-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

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 ((5.556*10^-10)*5*10^-2) = 5.27*10^-6

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

so, x = 5.27*10^-6 M

[OH-] = x = 5.27*10^-6 M

use:

pOH = -log [OH-]

= -log (5.27*10^-6)

= 5.2782

use:

PH = 14 - pOH

= 14 - 5.2782

= 8.7218

Answer: 8.72

5)when 60.0 mL of NaOH is added

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 60 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 60 mL = 6 mmol

We have:

mol(CH3COOH) = 5 mmol

mol(NaOH) = 6 mmol

5 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 50 + 60 = 110 mL

[OH-] = 1 mmol/110 mL = 0.0091 M

use:

pOH = -log [OH-]

= -log (9.091*10^-3)

= 2.0414

use:

PH = 14 - pOH

= 14 - 2.0414

= 11.9586

Answer: 11.96