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