Question

A 25.0mL sample of 0.35M HCOOH is titrated with 0.20 M KOH.
a) Determine the volume required to reach the equivalence point
b) the pH of the solution at the equivalence point. K_a=1.77x10^-4

Answer 1

a)

find the volume of KOH used to reach equivalence point

M(HCOOH)*V(HCOOH) =M(KOH)*V(KOH)

0.35 M *25.0 mL = 0.20 M *V(KOH)

V(KOH) = 43.75 mL

Answer: 43.75 mL

b)

we have:

Molarity of HCOOH = 0.35 M

Volume of HCOOH = 25 mL

Molarity of KOH = 0.2 M

Volume of KOH = 43.75 mL

mol of HCOOH = Molarity of HCOOH * Volume of HCOOH

mol of HCOOH = 0.35 M * 25 mL = 8.75 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.2 M * 43.75 mL = 8.75 mmol

We have:

mol of HCOOH = 8.75 mmol

mol of KOH = 8.75 mmol

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

HCOO- here is strong base

HCOO- formed = 8.75 mmol

Volume of Solution = 25 + 43.75 = 68.75 mL

Kb of HCOO- = Kw/Ka = 1*10^-14/1.77*10^-4 = 5.65*10^-11

concentration ofHCOO-,c = 8.75 mmol/68.75 mL = 0.1273M

HCOO- dissociates as

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

0.1273 0 0

0.1273-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.65*10^-11)*0.1273) = 2.682*10^-6

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

so, x = 2.682*10^-6 M

[OH-] = x = 2.682*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.682*10^-6)

= 5.5716

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.5716

= 8.43

Answer: 8.43