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Topic: NH4OH-NH4Cl problem  (Read 47267 times)

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tashkent

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NH4OH-NH4Cl problem
« on: February 25, 2005, 12:18:38 PM »
Hello everyone!

I’m having a big problem solving some questions regarding NH4OH-NH4Cl buffer system.  Here are the questions:

1)   What is the [H+] of a solution which is 0.1 M in NH4OH and 0.1 M in NH4Cl?

2)   If 1.0 L of the solution above is added to 250 mL of 0.1 M HCl, what is the resulting pH of the solution formed?

3)   If 1.0 L of the solution in #1 is added to 500 mL of 0.1 M HCl, what is the pH of the resulting solution?

4)   If 1.0 L of the solution in #1 is added to 1.0 L of 0.1 M HCl, what is the pH of the resulting solution?

5)   If 1.0 L of the solution containing only 0.1 M NH4OH is added to 1.0 L of 0.1 M HCl, what is the pH of the resulting solution?

Given : Kb NH4OH = 1.8 x 10-5

Here were my answers:

1)   NH4OH (0.1 M) = NH4+ (0.1 M) + OH- (0.1 M)
NH4Cl (0.1 M) = NH4+ (0.1 M) + Cl- (0.1 M)

Therefore, [NH4+] from NH4OH and NH4Cl is 0.2 M.

NH4+(0.2 M) + H2O = NH3(x) + H3O+(x)

Let x = [H3O+], [NH3]

[H3O+] [NH3]/[NH4+] = Ka

Ka = Kw/Kb, therefore = [H3O+] [NH3]/[NH4+] = Kw/Kb

X2/0.2 = 10-14/1.8 x 10-5

X = ?2 x 10-15/1.8 x 10-5 = 1.05 x 10-5

[H3O+] = 1.1 x 10-5



2)   HCl (0.025 mol) = H+ + Cl-

[H+] = 0.025 mol/1.25 L = 0.02 M [H+] from HCl

Total [H+] = 1.05 x 10-5 (from # 1) + 0.02 M
         =   0.02

pH= 1.7


3)   HCl (0.05 mol) = H+ + Cl-

[H+] = 0.05 mol/1.5 L = 0.033 M

Total [H+] = 1.05 x 10-5(from # 1) + 0.033 M
        =   0.0333 M

pH = 1.48


4)   HCl (0.1 mol) = H+ + Cl-

[H+] = 0.1 mol/2 L = 0.05 M

Total [H+] = 1.05 x 10-5 + 0.05
      =  0.05

pH = 1.3

5)   NH4OH(0.1 M) = NH4+(0.1 M) + OH-(0.1 M)

NH4+(0.1 M) + H2O = NH3(x) + H3O+(x)

Let x = [H3O+], [NH3]

[H3O+] [NH3]/[NH4+] = Ka

Ka = Kw/Kb, therefore = [H3O+] [NH3]/[NH4+] = Kw/Kb

X2/0.1 = 10-14/1.8 x 10-5

X = ?1 x 10-15/1.8 x 10-5 = 7.4 x 10-6

[H3O+] = 7.4 x 10-6 M


HCl (0.1 mol) = H+ + Cl-

[H+] from HCl = 0.1 mol/2 L = 0.05 M

Total [H+] = 7.4 x 10-6 + 0.05 M
         =   0.05

pH = 1.3


To my dismay, my prof says my computations were wrong.  He said that the [NH4+] from both NH4OH and NH4Cl in # 1 should not be additive, therefore [NH4+] should not be 0.2 M.

I would appreciate any help from you in solving this problem.  Thank you.

Sincerely,
Tashkent

Offline Borek

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Re:NH4OH-NH4Cl problem
« Reply #1 on: February 25, 2005, 09:29:14 PM »
Havn't checked it thoroughly, but most of the questions are the buffer questions, so you should use so called Henderson-Hasselbalch equation.

Your basic mistake is that you are ignoring the fact, that NH4OH is a weak base and is not dissociated 100%.

Check out my program BATE - while it won't help you understand how the calculations should be done, at least you will be able to check what the proper answer is and compare it with your results.
« Last Edit: February 25, 2005, 09:32:27 PM by Borek »
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