Question

A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with 0.100 M KOH. The Ka for the weak acid formic is 1.40 x 10-5. a. Determine the pH for the formic prior to its titration with KOH. b. Determine the pH of this solution at the ½ neutralization point of the titration. c. Identify the conjugate acid-base pair species at the ½ neutralization point.

Answer 1

a) Prior to titration, no. of moles of formic acid = (0.100 mol/L)(0.020 L) = 0.002 mol

                 HCOOH      ---->    H⁺    +    HCOO^-

Initial       0.002               0            0

change       -x                +x    +x

equilibrium(moles) 0.002-x           x              x

equilibrium(conc.)   (0.002-x)/0.020      x/0.020         x/0.020

acid dissociation constant, K_a is given by:

  

Assuming x to be very small compared to 0.002 and neglecting it, the above expression becomes:

solving for x, we get:

x = 2.4 x 10^-5

Therefore, [H⁺] = x/0.020 = (2.4 x 10^-5)/0.020 = 0.0012 M

pH = -log[H⁺] = -log(0.0012) = 2.92

b) At half the neutralization point, pH = pK_a

So, pH = -log(1.4 x 10^-5) = 4.85

c) HCOOH + KOH ----> HCOO^- K⁺ + H₂O

conjugate acid-base pairs are:

HCOOH, HCOO^-

and, OH^- , H₂O