Question

Calculate the pH at the equivalence point for the following titration: 0.100 M NaOH versus 1.60 g formic acid (HCO2H Ka = 1.8 x 10-4).

Answer 1

at equivalence point all the formic acid will be converted to formate.

moles of formic acid used = 1.6g/46g/mol = 0.035 moles

moles of formate formed = 0.035 moles

At equivalence point some of the formate will hydrolyse back to formic acid according to the following equilibrium.

HCOO- + H2O <==> HCOOH + OH-

Kb of the reaction = 10^-14/Ka = 5.55 *10^-11

	HCOO-	HCOOH	OH-
initial	0.035	0	0
change	-x	+x	+x
equilibrium	0.035-x	x	x

Kb = x*x/0.035-x

or, 5.55 *10^-11 = x^2/0.035-x

or, 1.94*10^-12 - 5.55x *10^-11- x^2 = 0

or, x = 1.39*10^-6 M

[OH-] = 1.39*10^-6

pOH = -log [OH-] = 5.85

pH = 14-pOH = 8.14