Question

Weak acid with strong base: 25 mL of 0.1 M HCOOH with 0.1 M NaOH

(pka = 3.74)

pH after adding following amounts of NaOH

18) 0 mL of NaOH

19) at half equivalence point

20) after adding 25 mL NaOH

Answer 1

use:

pKa = -log Ka

3.74 = -log Ka

Ka = 1.82*10^-4

18)when 0.0 mL of NaOH is added

HCOOH dissociates as:

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

0.1 0 0

0.1-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 ((1.82*10^-4)*0.1) = 4.266*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)

1.82*10^-4 = x^2/(0.1-x)

1.82*10^-5 - 1.82*10^-4 *x = x^2

x^2 + 1.82*10^-4 *x-1.82*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.82*10^-4

c = -1.82*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 = 7.282*10^-5

roots are :

x = 4.176*10^-3 and x = -4.358*10^-3

since x can't be negative, the possible value of x is

x = 4.176*10^-3

use:

pH = -log [H+]

= -log (4.176*10^-3)

= 2.3793

Answer: 2.38

19)

At half equivalence point,

pH = pKa

= 3.74

Answer: 3.74

20)

Given:

M(HCOOH) = 0.1 M

V(HCOOH) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.1 M * 25 mL = 2.5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCOOH) = 2.5 mmol

mol(NaOH) = 2.5 mmol

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

HCOO- here is strong base

HCOO- formed = 2.5 mmol

Volume of Solution = 25 + 25 = 50 mL

Kb of HCOO- = Kw/Ka = 1*10^-14/1.82*10^-4 = 5.495*10^-11

concentration ofHCOO-,c = 2.5 mmol/50 mL = 0.05M

HCOO- dissociates as

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

0.05 0 0

0.05-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 ((5.495*10^-11)*5*10^-2) = 1.658*10^-6

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

so, x = 1.658*10^-6 M

[OH-] = x = 1.658*10^-6 M

use:

pOH = -log [OH-]

= -log (1.658*10^-6)

= 5.7805

use:

PH = 14 - pOH

= 14 - 5.7805

= 8.2195

Answer: 8.22