Question

Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 25.0 mL of 0.200 M HA with 0.100 M NaOH. Ka = 2.0 x 10-5.

Answer 1

1)when 0.0 mL of NaOH is added

HA dissociates as:

HA -----> H+ + A-

0.2 0 0

0.2-x x x

Ka = [H+][A-]/[HA]

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 ((2*10^-5)*0.2) = 2*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)

2*10^-5 = x^2/(0.2-x)

4*10^-6 - 2*10^-5 *x = x^2

x^2 + 2*10^-5 *x-4*10^-6 = 0

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

a = 1

b = 2*10^-5

c = -4*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 = 1.6*10^-5

roots are :

x = 1.99*10^-3 and x = -2.01*10^-3

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

x = 1.99*10^-3

use:

pH = -log [H+]

= -log (1.99*10^-3)

= 2.7011

Answer: 2.70

2)when 10.0 mL of NaOH is added

Given:

M(HA) = 0.2 M

V(HA) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 10 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.2 M * 25 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 10 mL = 1 mmol

We have:

mol(HA) = 5 mmol

mol(NaOH) = 1 mmol

1 mmol of both will react

excess HA remaining = 4 mmol

Volume of Solution = 25 + 10 = 35 mL

[HA] = 4 mmol/35 mL = 0.1143M

[A-] = 1/35 = 0.0286M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 2*10^-5

pKa = - log (Ka)

= - log(2*10^-5)

= 4.699

use:

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

= 4.699+ log {2.857*10^-2/0.1143}

= 4.097

Answer: 4.10

3)when 25.0 mL of NaOH is added

Given:

M(HA) = 0.2 M

V(HA) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.2 M * 25 mL = 5 mmol

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

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

We have:

mol(HA) = 5 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess HA remaining = 2.5 mmol

Volume of Solution = 25 + 25 = 50 mL

[HA] = 2.5 mmol/50 mL = 0.05M

[A-] = 2.5/50 = 0.05M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 2*10^-5

pKa = - log (Ka)

= - log(2*10^-5)

= 4.699

use:

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

= 4.699+ log {5*10^-2/5*10^-2}

= 4.699

Answer: 4.70

4)when 50.0 mL of NaOH is added

Given:

M(HA) = 0.2 M

V(HA) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 50 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.2 M * 25 mL = 5 mmol

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

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

We have:

mol(HA) = 5 mmol

mol(NaOH) = 5 mmol

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

A- here is strong base

A- formed = 5 mmol

Volume of Solution = 25 + 50 = 75 mL

Kb of A- = Kw/Ka = 1*10^-14/2*10^-5 = 5*10^-10

concentration ofA-,c = 5 mmol/75 mL = 0.0667M

A- dissociates as

A- + H2O -----> HA + OH-

0.0667 0 0

0.0667-x x x

Kb = [HA][OH-]/[A-]

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*10^-10)*6.667*10^-2) = 5.774*10^-6

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

so, x = 5.774*10^-6 M

[OH-] = x = 5.774*10^-6 M

use:

pOH = -log [OH-]

= -log (5.774*10^-6)

= 5.2386

use:

PH = 14 - pOH

= 14 - 5.2386

= 8.7614

Answer: 8.76

5)when 60.0 mL of NaOH is added

Given:

M(HA) = 0.2 M

V(HA) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 60 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.2 M * 25 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 60 mL = 6 mmol

We have:

mol(HA) = 5 mmol

mol(NaOH) = 6 mmol

5 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 25 + 60 = 85 mL

[OH-] = 1 mmol/85 mL = 0.0118 M

use:

pOH = -log [OH-]

= -log (1.176*10^-2)

= 1.9294

use:

PH = 14 - pOH

= 14 - 1.9294

= 12.0706

Answer: 12.07