Question

What is the pH at the equivalence point when 50.00 mL of 0.112 M hydroxyacetic acid is titrated with 0.0580 M KOH? (Assume K_a = 1.47 ✕ 10⁻⁴, and K_w = 1.01 ✕ 10⁻¹⁴.)

Answer 1

For simplicity I am writing hydroxyacetic acid as HA

find the volume of KOH used to reach equivalence point

M(HA)*V(HA) =M(KOH)*V(KOH)

0.112 M *50.0 mL = 0.058M *V(KOH)

V(KOH) = 96.5517 mL

Given:

M(HA) = 0.112 M

V(HA) = 50 mL

M(KOH) = 0.058 M

V(KOH) = 96.5517 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.112 M * 50 mL = 5.6 mmol

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

mol(KOH) = 0.058 M * 96.5517 mL = 5.6 mmol

We have:

mol(HA) = 5.6 mmol

mol(KOH) = 5.6 mmol

5.6 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 5.6 mmol

Volume of Solution = 50 + 96.5517 = 146.5517 mL

Kb of A- = Kw/Ka = 1*10^-14/1.47*10^-4 = 6.803*10^-11

concentration ofA-,c = 5.6 mmol/146.5517 mL = 0.0382M

A- dissociates as

A- + H2O -----> HA + OH-

0.0382 0 0

0.0382-x x x

Kb = [HA][OH-]/[A-]

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 ((6.803*10^-11)*3.821*10^-2) = 1.612*10^-6

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

so, x = 1.612*10^-6 M

[OH-] = x = 1.612*10^-6 M

use:

pOH = -log [OH-]

= -log (1.612*10^-6)

= 5.79

use:

PH = 14 - pOH

= 14 - 5.79

= 8.21

pH = 8.21