Question

A 15.0 mL sample of 0.21 M nitrous acid (HNO₂), K_a = 4.6x10^-4) is titrated with 0.15 M NaOH. Calculate the pH after the addition of the following volumes of base:

a. 0.00 mL      

b. 10.00 mL    

c. 21.00 mL    

d. 25.00 mL

Answer 1

a)when 0.0 mL of NaOH is added

HNO2 dissociates as:

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

0.21 0 0

0.21-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 ((4.6*10^-4)*0.21) = 9.829*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)

4.6*10^-4 = x^2/(0.21-x)

9.66*10^-5 - 4.6*10^-4 *x = x^2

x^2 + 4.6*10^-4 *x-9.66*10^-5 = 0

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

a = 1

b = 4.6*10^-4

c = -9.66*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 = 3.866*10^-4

roots are :

x = 9.601*10^-3 and x = -1.006*10^-2

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

x = 9.601*10^-3

use:

pH = -log [H+]

= -log (9.601*10^-3)

= 2.0177

Answer: 2.02

b)when 10.0 mL of NaOH is added

Given:

M(HNO2) = 0.21 M

V(HNO2) = 15 mL

M(NaOH) = 0.15 M

V(NaOH) = 10 mL

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

mol(HNO2) = 0.21 M * 15 mL = 3.15 mmol

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

mol(NaOH) = 0.15 M * 10 mL = 1.5 mmol

We have:

mol(HNO2) = 3.15 mmol

mol(NaOH) = 1.5 mmol

1.5 mmol of both will react

excess HNO2 remaining = 1.65 mmol

Volume of Solution = 15 + 10 = 25 mL

[HNO2] = 1.65 mmol/25 mL = 0.066M

[NO2-] = 1.5/25 = 0.06M

They form acidic buffer

acid is HNO2

conjugate base is NO2-

Ka = 4.6*10^-4

pKa = - log (Ka)

= - log(4.6*10^-4)

= 3.337

use:

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

= 3.337+ log {6*10^-2/6.6*10^-2}

= 3.296

Answer: 3.30

c)when 21.0 mL of NaOH is added

Given:

M(HNO2) = 0.21 M

V(HNO2) = 15 mL

M(NaOH) = 0.15 M

V(NaOH) = 21 mL

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

mol(HNO2) = 0.21 M * 15 mL = 3.15 mmol

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

mol(NaOH) = 0.15 M * 21 mL = 3.15 mmol

We have:

mol(HNO2) = 3.15 mmol

mol(NaOH) = 3.15 mmol

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

NO2- here is strong base

NO2- formed = 3.15 mmol

Volume of Solution = 15 + 21 = 36 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.6*10^-4 = 2.174*10^-11

concentration ofNO2-,c = 3.15 mmol/36 mL = 0.0875M

NO2- dissociates as

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

0.0875 0 0

0.0875-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 ((2.174*10^-11)*8.75*10^-2) = 1.379*10^-6

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

so, x = 1.379*10^-6 M

[OH-] = x = 1.379*10^-6 M

use:

pOH = -log [OH-]

= -log (1.379*10^-6)

= 5.8604

use:

PH = 14 - pOH

= 14 - 5.8604

= 8.1396

Answer: 8.14

d)when 25.0 mL of NaOH is added

Given:

M(HNO2) = 0.21 M

V(HNO2) = 15 mL

M(NaOH) = 0.15 M

V(NaOH) = 25 mL

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

mol(HNO2) = 0.21 M * 15 mL = 3.15 mmol

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

mol(NaOH) = 0.15 M * 25 mL = 3.75 mmol

We have:

mol(HNO2) = 3.15 mmol

mol(NaOH) = 3.75 mmol

3.15 mmol of both will react

excess NaOH remaining = 0.6 mmol

Volume of Solution = 15 + 25 = 40 mL

[OH-] = 0.6 mmol/40 mL = 0.015 M

use:

pOH = -log [OH-]

= -log (1.5*10^-2)

= 1.8239

use:

PH = 14 - pOH

= 14 - 1.8239

= 12.1761

Answer: 12.18