Question

Determine the pH during the titration of 64.3 mL of 0.426 M nitrous acid (Ka = 4.5×10-4) by 0.426 M NaOH at the following points. (Assume the titration is done at 25 °C.)

(a) Before the addition of any NaOH

(b) After the addition of 14.0 mL of NaOH

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

(d) At the equivalence point

(e) After the addition of 96.5 mL of NaOH

Answer 1

a)when 0.0 mL of NaOH is added

HNO2 dissociates as:

HNO2 -----> H+ + NO2-

0.426 0 0

0.426-x x x

Ka = [H+][NO2-]/[HNO2]

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 ((4.5*10^-4)*0.426) = 1.385*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

4.5*10^-4 = x^2/(0.426-x)

1.917*10^-4 - 4.5*10^-4 *x = x^2

x^2 + 4.5*10^-4 *x-1.917*10^-4 = 0

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

a = 1

b = 4.5*10^-4

c = -1.917*10^-4

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 = 7.67*10^-4

roots are :

x = 1.362*10^-2 and x = -1.407*10^-2

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

x = 1.362*10^-2

use:

pH = -log [H+]

= -log (1.362*10^-2)

= 1.8657

Answer: 1.87

b)when 14.0 mL of NaOH is added

Given:

M(HNO2) = 0.426 M

V(HNO2) = 64.3 mL

M(NaOH) = 0.426 M

V(NaOH) = 14 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.426 M * 64.3 mL = 27.3918 mmol

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

mol(NaOH) = 0.426 M * 14 mL = 5.964 mmol

We have:

mol(HNO2) = 27.3918 mmol

mol(NaOH) = 5.964 mmol

5.964 mmol of both will react

excess HNO2 remaining = 21.4278 mmol

Volume of Solution = 64.3 + 14 = 78.3 mL

[HNO2] = 21.4278 mmol/78.3 mL = 0.2737M

[NO2-] = 5.964/78.3 = 0.0762M

They form acidic buffer

acid is HNO2

conjugate base is NO2-

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.347

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.347+ log {7.617*10^-2/0.2737}

= 2.791

Answer: 2.79

c)

At half equivalence point, pH = pKa

use:

pKa = -log Ka

= -log (4.5*10^-4)

= 3.3468

So, pH = 3.3468

Answer: 3.35

d)

find the volume of NaOH used to reach equivalence point

M(HNO2)*V(HNO2) =M(NaOH)*V(NaOH)

0.426 M *64.3 mL = 0.426M *V(NaOH)

V(NaOH) = 64.3 mL

Given:

M(HNO2) = 0.426 M

V(HNO2) = 64.3 mL

M(NaOH) = 0.426 M

V(NaOH) = 64.3 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.426 M * 64.3 mL = 27.3918 mmol

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

mol(NaOH) = 0.426 M * 64.3 mL = 27.3918 mmol

We have:

mol(HNO2) = 27.3918 mmol

mol(NaOH) = 27.3918 mmol

27.3918 mmol of both will react to form NO2- and H2O

NO2- here is strong base

NO2- formed = 27.3918 mmol

Volume of Solution = 64.3 + 64.3 = 128.6 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.5*10^-4 = 2.222*10^-11

concentration ofNO2-,c = 27.3918 mmol/128.6 mL = 0.213M

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.213 0 0

0.213-x x x

Kb = [HNO2][OH-]/[NO2-]

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 ((2.222*10^-11)*0.213) = 2.176*10^-6

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

so, x = 2.176*10^-6 M

[OH-] = x = 2.176*10^-6 M

use:

pOH = -log [OH-]

= -log (2.176*10^-6)

= 5.6624

use:

PH = 14 - pOH

= 14 - 5.6624

= 8.3376

Answer: 8.34

e)when 96.5 mL of NaOH is added

Given:

M(HNO2) = 0.426 M

V(HNO2) = 64.3 mL

M(NaOH) = 0.426 M

V(NaOH) = 96.5 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.426 M * 64.3 mL = 27.3918 mmol

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

mol(NaOH) = 0.426 M * 96.5 mL = 41.109 mmol

We have:

mol(HNO2) = 27.3918 mmol

mol(NaOH) = 41.109 mmol

27.3918 mmol of both will react

excess NaOH remaining = 13.7172 mmol

Volume of Solution = 64.3 + 96.5 = 160.8 mL

[OH-] = 13.7172 mmol/160.8 mL = 0.0853 M

use:

pOH = -log [OH-]

= -log (8.531*10^-2)

= 1.069

use:

PH = 14 - pOH

= 14 - 1.069

= 12.931

Answer: 12.93