Question

 a) A 50.0 mL solution of 0.200 M acetic acid (CH₃COOH), 50.0 mL of 0.200 M is titrated with 0.200 M NaOH. Determine the pH.of acetic acid before any NaOH is added. The Ka of CH₃COOH is 1.8 x 10^-5.

 b) Determine the pH of the solution at the equivalent point.

Answer 1

a)when 0.0 mL of NaOH is added

CH3COOH dissociates as:

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

0.2 0 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)*0.2) = 1.897*10^-3

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

so, x = 1.897*10^-3 M

use:

pH = -log [H+]

= -log (1.897*10^-3)

= 2.7218

Answer: 2.72

b)

find the volume of NaOH used to reach equivalence point

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

0.2 M *50.0 mL = 0.2M *V(NaOH)

V(NaOH) = 50 mL

Given:

M(CH3COOH) = 0.2 M

V(CH3COOH) = 50 mL

M(NaOH) = 0.2 M

V(NaOH) = 50 mL

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

mol(CH3COOH) = 0.2 M * 50 mL = 10 mmol

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

mol(NaOH) = 0.2 M * 50 mL = 10 mmol

We have:

mol(CH3COOH) = 10 mmol

mol(NaOH) = 10 mmol

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

CH3COO- here is strong base

CH3COO- formed = 10 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 = 10 mmol/100 mL = 0.1M

CH3COO- dissociates as

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

0.1 0 0

0.1-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)*0.1) = 7.454*10^-6

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

so, x = 7.454*10^-6 M

[OH-] = x = 7.454*10^-6 M

use:

pOH = -log [OH-]

= -log (7.454*10^-6)

= 5.1276

use:

PH = 14 - pOH

= 14 - 5.1276

= 8.8724

Answer: 8.87