Question

A 35.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. Express your answer using two decimal places.

A. 0.0 mL

B. 17.5 mL

C. 35.0 mL

D. 80.0 mL

Answer 1

1)when 0.0 mL of HNO3 is added

NH3 dissociates as:

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

0.15 0 0

0.15-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)*0.15) = 1.643*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/(0.15-x)

2.7*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-2.7*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -2.7*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 = 1.08*10^-5

roots are :

x = 1.634*10^-3 and x = -1.652*10^-3

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

x = 1.634*10^-3

So, [OH-] = x = 1.634*10^-3 M

use:

pOH = -log [OH-]

= -log (1.634*10^-3)

= 2.7867

use:

PH = 14 - pOH

= 14 - 2.7867

= 11.2133

2)when 17.5 mL of HNO3 is added

Given:

M(HNO3) = 0.15 M

V(HNO3) = 17.5 mL

M(NH3) = 0.15 M

V(NH3) = 35 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.15 M * 17.5 mL = 2.625 mmol

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

mol(NH3) = 0.15 M * 35 mL = 5.25 mmol

We have:

mol(HNO3) = 2.625 mmol

mol(NH3) = 5.25 mmol

2.625 mmol of both will react

excess NH3 remaining = 2.625 mmol

Volume of Solution = 17.5 + 35 = 52.5 mL

[NH3] = 2.625 mmol/52.5 mL = 0.05 M

[NH4+] = 2.625 mmol/52.5 mL = 0.05 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 {5*10^-2/5*10^-2}

= 4.745

use:

PH = 14 - pOH

= 14 - 4.7447

= 9.2553

3)when 35.0 mL of HNO3 is added

Given:

M(HNO3) = 0.15 M

V(HNO3) = 35 mL

M(NH3) = 0.15 M

V(NH3) = 35 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NH3) = 0.15 M * 35 mL = 5.25 mmol

We have:

mol(HNO3) = 5.25 mmol

mol(NH3) = 5.25 mmol

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

NH4+ here is strong acid

NH4+ formed = 5.25 mmol

Volume of Solution = 35 + 35 = 70 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 5.25 mmol/70 mL = 0.075 M

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

7.5*10^-2 0 0

7.5*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 ((5.556*10^-10)*7.5*10^-2) = 6.455*10^-6

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

so, x = 6.455*10^-6 M

[H+] = x = 6.455*10^-6 M

use:

pH = -log [H+]

= -log (6.455*10^-6)

= 5.1901

4)when 80.0 mL of HNO3 is added

Given:

M(HNO3) = 0.15 M

V(HNO3) = 80 mL

M(NH3) = 0.15 M

V(NH3) = 35 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.15 M * 80 mL = 12 mmol

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

mol(NH3) = 0.15 M * 35 mL = 5.25 mmol

We have:

mol(HNO3) = 12 mmol

mol(NH3) = 5.25 mmol

5.25 mmol of both will react

excess HNO3 remaining = 6.75 mmol

Volume of Solution = 80 + 35 = 115 mL

[H+] = 6.75 mmol/115 mL = 0.0587 M

use:

pH = -log [H+]

= -log (5.87*10^-2)

= 1.2314