Question

A titration is performed on 50mL of a 0.10 M solution of HN3 (Ka+1.9x10-5) with 0.10 M NaOH as the titrant. (a) What is the initial pH of the HN3 solution? (b) Calculate the pH of the solution after 25 ml of NaOH have been added (c) How many mL of NaOH are required to reach the equivalence point and what is the pH of the solution at the equivalence point.

Answer 1

a)

HN3 dissociates as:

HN3 -----> H+ + N3-

0.1 0 0

0.1-x x x

Ka = [H+][N3-]/[HN3]

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 ((1.9*10^-5)*0.1) = 1.378*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)

1.9*10^-5 = x^2/(0.1-x)

1.9*10^-6 - 1.9*10^-5 *x = x^2

x^2 + 1.9*10^-5 *x-1.9*10^-6 = 0

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

a = 1

b = 1.9*10^-5

c = -1.9*10^-6

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 = 7.6*10^-6

roots are :

x = 1.369*10^-3 and x = -1.388*10^-3

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

x = 1.369*10^-3

use:

pH = -log [H+]

= -log (1.369*10^-3)

= 2.86

b)

Given:

M(HN3) = 0.1 M

V(HN3) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(HN3) = M(HN3) * V(HN3)

mol(HN3) = 0.1 M * 50 mL = 5 mmol

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

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

We have:

mol(HN3) = 5 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess HN3 remaining = 2.5 mmol

Volume of Solution = 50 + 25 = 75 mL

[HN3] = 2.5 mmol/75 mL = 0.0333M

[N3-] = 2.5/75 = 0.0333M

They form acidic buffer

acid is HN3

conjugate base is N3-

Ka = 1.9*10^-5

pKa = - log (Ka)

= - log(1.9*10^-5)

= 4.721

use:

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

= 4.721+ log {3.333*10^-2/3.333*10^-2}

= 4.721

Answer: 4.72

c)

use:

pKa = -log Ka

3.4 = -log Ka

Ka = 3.981*10^-4

find the volume of NaOH used to reach equivalence point

M(HN3)*V(HN3) =M(NaOH)*V(NaOH)

0.1 M *50.0 mL = 0.1M *V(NaOH)

V(NaOH) = 50 mL

Given:

M(HN3) = 0.1 M

V(HN3) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 50 mL

mol(HN3) = M(HN3) * V(HN3)

mol(HN3) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 50 mL = 5 mmol

We have:

mol(HN3) = 5 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react to form N3- and H2O

N3- here is strong base

N3- formed = 5 mmol

Volume of Solution = 50 + 50 = 100 mL

Kb of N3- = Kw/Ka = 1*10^-14/3.981*10^-4 = 2.512*10^-11

concentration ofN3-,c = 5 mmol/100 mL = 0.05M

N3- dissociates as

N3- + H2O -----> HN3 + OH-

0.05 0 0

0.05-x x x

Kb = [HN3][OH-]/[N3-]

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 ((2.512*10^-11)*5*10^-2) = 1.121*10^-6

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

so, x = 1.121*10^-6 M

[OH-] = x = 1.121*10^-6 M

use:

pOH = -log [OH-]

= -log (1.121*10^-6)

= 5.9505

use:

PH = 14 - pOH

= 14 - 5.9505

= 8.0495

Answer:

50 mL

8.05