Question

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL 21) of aqueous formic acid requires 29.80 mL of 0.0567 M NaOH? Ka =1.8 × 10-4 for formic acid.

A) 2.46 B) 8.12 C) 11.54 D) 5.88

Answer 1

given equivalence point

we know that

at equivalence point

moles of acid = moles of base added

also

we know that

moles = molarity x volume

so

Ma x Va = Mb x Vb

so

using the given values

we get

Ma x 25 = 0.0567 x 29.80

Ma = 0.0676

so

the molarity of acid is 0.0676

now

moles of base added = molarity x volume (L)

so

moles of NaOH added = 0.0567 x 29.80 x 10-3

moles of NaOH added = 1.69 x 10-3

now

the reaction is

HCOOH + NaOH ---> HCOONa + H20

at equivalnce point

equal moles of HCOOH and NaOH reacts to give HCOONa

so

moles of HCOONa formed = 1.69 x 10-3

total volume = 25 + 29.80

total volume = 54.8 ml

so

conc of HCOONa = moles / volume (L)

conc of HCOONa = 1.69 x 10-3 / 54.8 x 10-3

conc of HCOONa = 0.03084

also

Kb = kw / Ka

so

Kb = 10-14 / 1.8 x 10-4

Kb = 5.555 x 10-11

now

HCOO- + H20 ---> HCOOH + OH-

Kb = [HCOOH] [OH-] / [HC00-]

we know that

[HCOOH] = [OH-]

so

Kb = [OH-]^2 / [HC00-]

[OH-]2 = kb [HC00-]

[OH-] = sqrt ( Kb x [HC00-] )

so

[OH-] = sqrt ( 5.555 x 10-11 x 0.03084 )

[OH-] = 1.3089 x 10-6

we know that

pOH = -log [OH-]

so

pOH = -log 1.3089 x 10-6

pOH = 5.88

now

pH = 14 - pOH

pH = 14 - 5.88

pH = 8.12

so

the pH at equivalence point is 8.12

so

the answer is option B) 8.12