Question

Determine the pH during the titration of 25.5 mL of 0.455 M formic acid (K_a = 1.8×10^-4) by 0.477 M KOH at the following points.

(a) Before the addition of any KOH

(b) After the addition of 5.90 mL of KOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 36.5 mL of KOH

Answer 1

1)when 0.0 mL of KOH is added
HCOOH dissociates as:

HCOOH          ----->     H+   + HCOO-
0.455                 0         0
0.455-x               x         x


Ka = [H+][HCOO-]/[HCOOH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.8*10^-4)*0.455) = 9.05*10^-3

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.8*10^-4 = x^2/(0.455-x)
8.19*10^-5 - 1.8*10^-4 *x = x^2
x^2 + 1.8*10^-4 *x-8.19*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-4
c = -8.19*10^-5

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.276*10^-4

roots are :
x = 8.96*10^-3 and x = -9.14*10^-3

since x can't be negative, the possible value of x is
x = 8.96*10^-3

use:
pH = -log [H+]
= -log (8.96*10^-3)
= 2.0477
Answer: 2.05

2)when 5.9 mL of KOH is added
Given:
M(HCOOH) = 0.455 M
V(HCOOH) = 25.5 mL
M(KOH) = 0.477 M
V(KOH) = 5.9 mL


mol(HCOOH) = M(HCOOH) * V(HCOOH)
mol(HCOOH) = 0.455 M * 25.5 mL = 11.6025 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.477 M * 5.9 mL = 2.8143 mmol


We have:
mol(HCOOH) = 11.6025 mmol
mol(KOH) = 2.8143 mmol

2.8143 mmol of both will react

excess HCOOH remaining = 8.7882 mmol
Volume of Solution = 25.5 + 5.9 = 31.4 mL
[HCOOH] = 8.7882 mmol/31.4 mL = 0.2799M

[HCOO-] = 2.8143/31.4 = 0.0896M

They form acidic buffer
acid is HCOOH
conjugate base is HCOO-


Ka = 1.8*10^-4

pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {8.963*10^-2/0.2799}
= 3.25


Answer: 3.25

3)
At half equivalence point, pH = pKa

use:
pKa = -log Ka
= -log (1.8*10^-4)
= 3.7447

So, pH = 3.7447
Answer: 3.74

4)
find the volume of KOH used to reach equivalence point
M(HCOOH)*V(HCOOH) =M(KOH)*V(KOH)
0.455 M *25.5 mL = 0.477M *V(KOH)
V(KOH) = 24.3239 mL
Given:
M(HCOOH) = 0.455 M
V(HCOOH) = 25.5 mL
M(KOH) = 0.477 M
V(KOH) = 24.3239 mL


mol(HCOOH) = M(HCOOH) * V(HCOOH)
mol(HCOOH) = 0.455 M * 25.5 mL = 11.6025 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.477 M * 24.3239 mL = 11.6025 mmol


We have:
mol(HCOOH) = 11.6025 mmol
mol(KOH) = 11.6025 mmol

11.6025 mmol of both will react to form HCOO- and H2O

HCOO- here is strong base
HCOO- formed = 11.6025 mmol
Volume of Solution = 25.5 + 24.3239 = 49.8239 mL
Kb of HCOO- = Kw/Ka = 1*10^-14/1.8*10^-4 = 5.556*10^-11
concentration ofHCOO-,c = 11.6025 mmol/49.8239 mL = 0.2329M

HCOO- dissociates as

HCOO-        + H2O   ----->     HCOOH +   OH-
0.2329                        0         0
0.2329-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)*0.2329) = 3.597*10^-6

since c is much greater than x, our assumption is correct
so, x = 3.597*10^-6 M



[OH-] = x = 3.597*10^-6 M

use:
pOH = -log [OH-]
= -log (3.597*10^-6)
= 5.4441


use:
PH = 14 - pOH
= 14 - 5.4441
= 8.5559
Answer: 8.56

5)when 36.5 mL of KOH is added
Given:
M(HCOOH) = 0.455 M
V(HCOOH) = 25.5 mL
M(KOH) = 0.477 M
V(KOH) = 36.5 mL


mol(HCOOH) = M(HCOOH) * V(HCOOH)
mol(HCOOH) = 0.455 M * 25.5 mL = 11.6025 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.477 M * 36.5 mL = 17.4105 mmol


We have:
mol(HCOOH) = 11.6025 mmol
mol(KOH) = 17.4105 mmol

11.6025 mmol of both will react

excess KOH remaining = 5.808 mmol
Volume of Solution = 25.5 + 36.5 = 62 mL
[OH-] = 5.808 mmol/62 mL = 0.0937 M

use:
pOH = -log [OH-]
= -log (9.368*10^-2)
= 1.0284


use:
PH = 14 - pOH
= 14 - 1.0284
= 12.9716
Answer: 12.97