Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 40 mL of 0.18 M butanoic acid(aq) with 0.28 M NaOH(aq). The K_a of CH₃CH₂CH₂COOH is 1.5 x 10^-5.

Answer 1

find the volume of NaOH used to reach equivalence point

M(CH3CH2CH2COOH)*V(CH3CH2CH2COOH) =M(NaOH)*V(NaOH)

0.18 M *40.0 mL = 0.28M *V(NaOH)

V(NaOH) = 25.7143 mL

Given:

M(CH3CH2CH2COOH) = 0.18 M

V(CH3CH2CH2COOH) = 40 mL

M(NaOH) = 0.28 M

V(NaOH) = 25.7143 mL

mol(CH3CH2CH2COOH) = M(CH3CH2CH2COOH) * V(CH3CH2CH2COOH)

mol(CH3CH2CH2COOH) = 0.18 M * 40 mL = 7.2 mmol

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

mol(NaOH) = 0.28 M * 25.7143 mL = 7.2 mmol

We have:

mol(CH3CH2CH2COOH) = 7.2 mmol

mol(NaOH) = 7.2 mmol

7.2 mmol of both will react to form CH3CH2CH2COO- and H2O

CH3CH2CH2COO- here is strong base

CH3CH2CH2COO- formed = 7.2 mmol

Volume of Solution = 40 + 25.7143 = 65.7143 mL

Kb of CH3CH2CH2COO- = Kw/Ka = 1*10^-14/1.5*10^-5 = 6.667*10^-10

concentration ofCH3CH2CH2COO-,c = 7.2 mmol/65.7143 mL = 0.1096M

CH3CH2CH2COO- dissociates as

CH3CH2CH2COO- + H2O -----> CH3CH2CH2COOH + OH-

0.1096 0 0

0.1096-x x x

Kb = [CH3CH2CH2COOH][OH-]/[CH3CH2CH2COO-]

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 ((6.667*10^-10)*0.1096) = 8.547*10^-6

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

so, x = 8.547*10^-6 M

[OH-] = x = 8.547*10^-6 M

use:

pOH = -log [OH-]

= -log (8.547*10^-6)

= 5.0682

use:

PH = 14 - pOH

= 14 - 5.0682

= 8.9318

Answer: 8.93