Question

What is the pH at the equivalence point of the titration of 0.150 L of 0.500 M formic acid with 30.0 mL of 2.50 M sodium hydroxide. Ka = 1.8 X10^-4

Answer 1

Given:

M(HCOOH) = 0.5 M

V(HCOOH) = 150 mL

M(NaOH) = 2.5 M

V(NaOH) = 30 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.5 M * 150 mL = 75 mmol

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

mol(NaOH) = 2.5 M * 30 mL = 75 mmol

We have:

mol(HCOOH) = 75 mmol

mol(NaOH) = 75 mmol

75 mmol of both will react to form HCOO- and H2O

HCOO- here is strong base

HCOO- formed = 75 mmol

Volume of Solution = 150 + 30 = 180 mL

Kb of HCOO- = Kw/Ka = 1*10^-14/1.8*10^-4 = 5.556*10^-11

concentration ofHCOO-,c = 75 mmol/180 mL = 0.4167M

HCOO- dissociates as

HCOO- + H2O -----> HCOOH + OH-

0.4167 0 0

0.4167-x x x

Kb = [HCOOH][OH-]/[HCOO-]

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^-11)*0.4167) = 4.811*10^-6

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

so, x = 4.811*10^-6 M

[OH-] = x = 4.811*10^-6 M

use:

pOH = -log [OH-]

= -log (4.811*10^-6)

= 5.3177

use:

PH = 14 - pOH

= 14 - 5.3177

= 8.6823

Answer: 8.68