Question

Calculate the pH of a solution produced during the titration of 50.00 mL of 2.0 M acetic acid with 1.0 M potassium hydroxide and the following volumes of base added:

a) 0.00 mL

b) 25.0 mL

c) 50.0 mL

d) 100.0 mL

e) 125.0 mL

Answer 1

a)

CH3COOH dissociates as:

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

2 0 0

2-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)*2) = 6*10^-3

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

so, x = 6*10^-3 M

so.[H+] = x = 6*10^-3 M

use:

pH = -log [H+]

= -log (6*10^-3)

= 2.22

pH is 2.22

b)

Given:

M(CH3COOH) = 2 M

V(CH3COOH) = 50 mL

M(KOH) = 1 M

V(KOH) = 25 mL

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

mol(CH3COOH) = 2 M * 50 mL = 100 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1 M * 25 mL = 25 mmol

We have:

mol(CH3COOH) = 100 mmol

mol(KOH) = 25 mmol

25 mmol of both will react

excess CH3COOH remaining = 75 mmol

Volume of Solution = 50 + 25 = 75 mL

[CH3COOH] = 75 mmol/75 mL = 1M

[CH3COO-] = 25/75 = 0.3333M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

use:

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

= 4.7447+ log {0.3333/1}

= 4.2676

pH = 4.27

c)

Given:

M(CH3COOH) = 2 M

V(CH3COOH) = 50 mL

M(KOH) = 1 M

V(KOH) = 50 mL

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

mol(CH3COOH) = 2 M * 50 mL = 100 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1 M * 50 mL = 50 mmol

We have:

mol(CH3COOH) = 100 mmol

mol(KOH) = 50 mmol

50 mmol of both will react

excess CH3COOH remaining = 50 mmol

Volume of Solution = 50 + 50 = 100 mL

[CH3COOH] = 50 mmol/100 mL = 0.5M

[CH3COO-] = 50/100 = 0.5M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

use:

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

= 4.7447+ log {0.5/0.5}

= 4.7447

pH = 4.75

d)

Given:

M(CH3COOH) = 2 M

V(CH3COOH) = 50 mL

M(KOH) = 1 M

V(KOH) = 100 mL

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

mol(CH3COOH) = 2 M * 50 mL = 100 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1 M * 100 mL = 100 mmol

We have:

mol(CH3COOH) = 100 mmol

mol(KOH) = 100 mmol

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

CH3COO- here is strong base

CH3COO- formed = 100 mmol

Volume of Solution = 50 + 100 = 150 mL

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

concentration ofCH3COO-,c = 100 mmol/150 mL = 0.6667M

CH3COO- dissociates as

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

0.6667 0 0

0.6667-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)*0.6667) = 1.925*10^-5

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

so, x = 1.925*10^-5 M

[OH-] = x = 1.925*10^-5 M

use:

pOH = -log [OH-]

= -log (1.925*10^-5)

= 4.7157

use:

PH = 14 - pOH

= 14 - 4.7157

= 9.2843

pH = 9.28

e)

Given:

M(CH3COOH) = 2 M

V(CH3COOH) = 50 mL

M(KOH) = 1 M

V(KOH) = 125 mL

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

mol(CH3COOH) = 2 M * 50 mL = 100 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1 M * 125 mL = 125 mmol

We have:

mol(CH3COOH) = 100 mmol

mol(KOH) = 125 mmol

100 mmol of both will react

excess KOH remaining = 25 mmol

Volume of Solution = 50 + 125 = 175 mL

[OH-] = 25 mmol/175 mL = 0.1429 M

use:

pOH = -log [OH-]

= -log (0.1429)

= 0.8451

use:

PH = 14 - pOH

= 14 - 0.8451

= 13.1549

pH = 13.2