[twistedesoterix]

Problem:  What are the final hydrogen ion concentration and pH of a solution obtained by mixing 200 mL of 0.4 M aqueous NH₃ with 300 mL of 0.2 M HCL? (K_B = 1.8 X 10^-5)

My first attempt at solving: 
@ t₀ I have 100% in the acid form (HCl)

mol HCl present = 0.3 L x 0.2 M = 0.06 mol HCl

mol HCl present = mol NH₃ required
0.06 mol HCl = (x amount of NH₃ in Liters) (0.4 M)
x L = 0.15 L = 150L of NH_3(aq)  :larrow:that doesn't make sense to me...

My 2nd attempt at solving:

pOH = pK_b + log (mol_HCl / mol_NH3)
pOH = 4.74473 + log (.08/.06) = 4.86967

pH = 14 - pOH = 14 - 4.86967 = 9.13033

[H⁺] = 7.41E-10 M

 
The answer key states that the
[H⁺] = 1.66E-9 M and pH = 8.78

What am I doing wrong?  Thank you in advance!!!

[Hunter2]

The first attempt is correct. But it is 150 ml ammonia.
So 50 ml ammonia is left from the 200 ml So you can calculate the mole s of left ammonia and the ammonium ions

Use Henderson Hasselbalch.

[Borek]

Two questions that you have to start with:

1. What is the reaction that takes place?

2. If it is a buffer solution - what is the base and what is its conjugate acid?

[twistedesoterix]

Two questions that you have to start with:

1. What is the reaction that takes place?
   

2. If it is a buffer solution - what is the base and what is its conjugate acid?
     

Am I wrong in thinking that this is an Acid/Base titration? Maybe not a buffer. Just an acid/base titration. I have an NH4OH as the base and the HCl as my acid.
Am I thinking correctly or am I all wrong?

[Borek]

You are wrong.

It is a titration that creates a buffer solution.

Please answer questions that I already asked.

[twistedesoterix]

HCl + NH₄OH  NH₄Cl + HOH
HCl - conj acid 1
NH₄OH - conj base 2
NH₄Cl - conj base 1
HOH - conj. acid 2

Is that right?

[Borek]

HCl + NH₄OH  NH₄Cl + HOH
HCl - conj acid 1
NH₄OH - conj base 2
NH₄Cl - conj base 1
HOH - conj. acid 2

Is that right?

No.

First of all - go net ionic. There is no such thing as NH₄Cl or HCl in the solution.

[twistedesoterix]

HCL + NH₃  NH₄⁺ + Cl^-

HCL - conj acid 1
NH3 - conj base 2
CL - conj base 1
NH4+ conj acid2

[Borek]

Much better now, although still not perfect. HCl is so strong an acid, you can assume it is 100% dissociated all the time and concentration of HCl can be safely ignored.

Do you see now conjugate acid/base pair responsible for pH? Can you calculate concentrations of the base and acid and plug them into Henderson-Hasselbalch equation?

[twistedesoterix]

I know that the H+ from the HCL is what's gonna give me the [H+]_final, which will give me the pH. Aqueous NH3 will yield OH^- (i'm not sure about this) which will give me the final [OH-] & pOH?

I really have no idea what to do next. All the examples we were given by teacher and on the book asks how much x M of "base" we need to neutralize xL of xM "acid" and i took chemistry back in 2005.

[Borek]

I know that the H+ from the HCL is what's gonna give me the [H+]_final, which will give me the pH. Aqueous NH3 will yield OH^- (i'm not sure about this) which will give me the final [OH-] & pOH?

No & no.

See if this helps: http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q2

It is a slightly different problem, but related.