Question

A solution of 100. ml of .500 M Acetic Acid is titrated with .500 M sodium hydroxide. The Ka of acetic acid is 1.8*10^-5. Find the pH values at the given stages: a) before the addition of any NaOH. B) After 25.0 mL of NaOH added. C) At the equivalence point.

Answer 1

A)when 0.0 mL of NaOH is added

CH3COOH dissociates as:

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

0.5 0 0

0.5-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)*0.5) = 3*10^-3

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

so, x = 3*10^-3 M

use:

pH = -log [H+]

= -log (3*10^-3)

= 2.5229

Answer: 2.52

B)when 25.0 mL of NaOH is added

Given:

M(CH3COOH) = 0.5 M

V(CH3COOH) = 100 mL

M(NaOH) = 0.5 M

V(NaOH) = 25 mL

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

mol(CH3COOH) = 0.5 M * 100 mL = 50 mmol

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

mol(NaOH) = 0.5 M * 25 mL = 12.5 mmol

We have:

mol(CH3COOH) = 50 mmol

mol(NaOH) = 12.5 mmol

12.5 mmol of both will react

excess CH3COOH remaining = 37.5 mmol

Volume of Solution = 100 + 25 = 125 mL

[CH3COOH] = 37.5 mmol/125 mL = 0.3M

[CH3COO-] = 12.5/125 = 0.1M

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 {0.1/0.3}

= 4.268

Answer: 4.27

C)

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.5 M *100.0 mL = 0.5M *V(NaOH)

V(NaOH) = 100 mL

Given:

M(CH3COOH) = 0.5 M

V(CH3COOH) = 100 mL

M(NaOH) = 0.5 M

V(NaOH) = 100 mL

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

mol(CH3COOH) = 0.5 M * 100 mL = 50 mmol

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

mol(NaOH) = 0.5 M * 100 mL = 50 mmol

We have:

mol(CH3COOH) = 50 mmol

mol(NaOH) = 50 mmol

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

CH3COO- here is strong base

CH3COO- formed = 50 mmol

Volume of Solution = 100 + 100 = 200 mL

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

concentration ofCH3COO-,c = 50 mmol/200 mL = 0.25M

CH3COO- dissociates as

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

0.25 0 0

0.25-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)*0.25) = 1.179*10^-5

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

so, x = 1.179*10^-5 M

[OH-] = x = 1.179*10^-5 M

use:

pOH = -log [OH-]

= -log (1.179*10^-5)

= 4.9287

use:

PH = 14 - pOH

= 14 - 4.9287

= 9.0713

Answer: 9.07