Question

In a titration of a 250mL solution of 0.05M ammonia (NH3) (Kb=1.8 * 10^-5) with a concentrated (0.2M) solution of HCL, calculate the following:

1. What is the initial pH before adding HCL?

2. What is the pH of the system after adding 20mL of the 0.2 M HCL?

3. What is the pH at equivalence?

*I have the answers, I really need to know how to do the work to get to the answers.*

Answer 1

1)

NH3 dissociates as:

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

5*10^-2 0 0

5*10^-2-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)*5*10^-2) = 9.487*10^-4

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/(5*10^-2-x)

9*10^-7 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-9*10^-7 = 0

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

a = 1

b = 1.8*10^-5

c = -9*10^-7

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 = 3.6*10^-6

roots are :

x = 9.397*10^-4 and x = -9.577*10^-4

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

x = 9.397*10^-4

So, [OH-] = x = 9.397*10^-4 M

use:

pOH = -log [OH-]

= -log (9.397*10^-4)

= 3.027

use:

PH = 14 - pOH

= 14 - 3.027

= 10.973

Answer: 10.97

2)

Given:

M(HCl) = 0.2 M

V(HCl) = 20 mL

M(NH3) = 0.05 M

V(NH3) = 250 mL

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

mol(HCl) = 0.2 M * 20 mL = 4 mmol

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

mol(NH3) = 0.05 M * 250 mL = 12.5 mmol

We have:

mol(HCl) = 4 mmol

mol(NH3) = 12.5 mmol

4 mmol of both will react

excess NH3 remaining = 8.5 mmol

Volume of Solution = 20 + 250 = 270 mL

[NH3] = 8.5 mmol/270 mL = 0.0315 M

[NH4+] = 4 mmol/270 mL = 0.0148 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 {1.481*10^-2/3.148*10^-2}

= 4.417

use:

PH = 14 - pOH

= 14 - 4.4174

= 9.5826

Answer: 9.58

3)

find the volume of HCl used to reach equivalence point

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

0.05 M *250.0 mL = 0.2M *V(HCl)

V(HCl) = 62.5 mL

Given:

M(HCl) = 0.2 M

V(HCl) = 62.5 mL

M(NH3) = 0.05 M

V(NH3) = 250 mL

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

mol(HCl) = 0.2 M * 62.5 mL = 12.5 mmol

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

mol(NH3) = 0.05 M * 250 mL = 12.5 mmol

We have:

mol(HCl) = 12.5 mmol

mol(NH3) = 12.5 mmol

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

NH4+ here is strong acid

NH4+ formed = 12.5 mmol

Volume of Solution = 62.5 + 250 = 312.5 mL

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

concentration ofNH4+,c = 12.5 mmol/312.5 mL = 0.04 M

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

4*10^-2 0 0

4*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)*4*10^-2) = 4.714*10^-6

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

so, x = 4.714*10^-6 M

[H+] = x = 4.714*10^-6 M

use:

pH = -log [H+]

= -log (4.714*10^-6)

= 5.3266

Answer: 5.33