Question

25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 50 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.

Answer 1

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.05 M

V(NaOH) = 50 mL

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

mol(CH3COOH) = 0.1 M * 25 mL = 2.5 mmol

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

mol(NaOH) = 0.05 M * 50 mL = 2.5 mmol

We have:

mol(CH3COOH) = 2.5 mmol

mol(NaOH) = 2.5 mmol

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

CH3COO- here is strong base

CH3COO- formed = 2.5 mmol

Volume of Solution = 25 + 50 = 75 mL

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

concentration ofCH3COO-,c = 2.5 mmol/75 mL = 0.0333M

CH3COO- dissociates as

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

0.0333 0 0

0.0333-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)*3.333*10^-2) = 4.303*10^-6

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

so, x = 4.303*10^-6 M

[OH-] = x = 4.303*10^-6 M

use:

pOH = -log [OH-]

= -log (4.303*10^-6)

= 5.3662

use:

PH = 14 - pOH

= 14 - 5.3662

= 8.6338

Answer: 8.63