Question

25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M HCl. Calculate the pH of the NH3 solution at the following points during the titration: (Kb= 1.8 x 10^-5) A. Prior to the addition of any HCl. B: After the addition of 10.5 mL of a 0.100 M HCl. C: At the equivilance point. D: After the addition of 3 mL of 0.100 M HCl. Show your work.

Answer 1

A)

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.1 0 0

0.1-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-5)*0.1) = 1.342*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

1.8*10^-5 = x^2/(0.1-x)

1.8*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

So, [OH-] = x = 1.333*10^-3 M

use:

pOH = -log [OH-]

= -log (1.333*10^-3)

= 2.8753

use:

PH = 14 - pOH

= 14 - 2.8753

= 11.12

B)

Given:

M(HCl) = 0.1 M

V(HCl) = 10.5 mL

M(NH3) = 0.1 M

V(NH3) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 10.5 mL = 1.05 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCl) = 1.05 mmol

mol(NH3) = 2.5 mmol

1.05 mmol of both will react

excess NH3 remaining = 1.45 mmol

Volume of Solution = 10.5 + 25 = 35.5 mL

[NH3] = 1.45 mmol/35.5 mL = 0.0408 M

[NH4+] = 1.05 mmol/35.5 mL = 0.0296 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {2.958*10^-2/4.085*10^-2}

= 4.605

use:

PH = 14 - pOH

= 14 - 4.6045

= 9.40

C)

use:

pKb = -log Kb

4.75= -log Kb

Kb = 1.778*10^-5

find the volume of HCl used to reach equivalence point

M(NH3)*V(NH3) =M(HCl)*V(HCl)

0.1 M *25.0 mL = 0.1M *V(HCl)

V(HCl) = 25 mL

Given:

M(HCl) = 0.1 M

V(HCl) = 25 mL

M(NH3) = 0.1 M

V(NH3) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

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

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCl) = 2.5 mmol

mol(NH3) = 2.5 mmol

2.5 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 2.5 mmol

Volume of Solution = 25 + 25 = 50 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.778279410038923E-5 = 5.623*10^-10

concentration ofNH4+,c = 2.5 mmol/50 mL = 0.05 M

NH4+ + H2O -----> NH3 + H+

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

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 ((5.623*10^-10)*5*10^-2) = 5.303*10^-6

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

so, x = 5.303*10^-6 M

[H+] = x = 5.303*10^-6 M

use:

pH = -log [H+]

= -log (5.303*10^-6)

= 5.2755

Answer: 5.28

D)

Given:

M(HCl) = 0.1 M

V(HCl) = 3 mL

M(NH3) = 0.1 M

V(NH3) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 3 mL = 0.3 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCl) = 0.3 mmol

mol(NH3) = 2.5 mmol

0.3 mmol of both will react

excess NH3 remaining = 2.2 mmol

Volume of Solution = 3 + 25 = 28 mL

[NH3] = 2.2 mmol/28 mL = 0.0786 M

[NH4+] = 0.3 mmol/28 mL = 0.0107 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {1.071*10^-2/7.857*10^-2}

= 3.879

use:

PH = 14 - pOH

= 14 - 3.8794

= 10.1206

Answer: 10.12