# Finding PH of solution

1. Apr 20, 2014

### Ritzycat

1. The problem statement, all variables and given/known data
20.0 mL of .200 M hydrochloric acid are mixed with 20.0 mL of .300M ammonium hydroxide. What is the pH?

Also says that if ammonium hydroxide is in excess it immediately breaks down into ammonia and water.

3. The attempt at a solution
I know the NH4OH is in excess, there are .006mol of NH4OH, .004 mol of HCl.

I'm not entirely too sure how to write the equation for this one.

HCl + NH4OH -> NH4Cl + H2O
HCl + NH3 + H2O -> NH4Cl + H2O
HCL + NH3 -> NH4Cl

Are any of these right to help me solve the equation? I don't know how to write the equation so I can find the concentration of the hydronium ions, thus finding the pH.

2. Apr 20, 2014

### Dr Transport

Hydrochloric acid is a strong acid and ammonium hydroxide is a strong base. You need to calculate the amount of hydronium ions and hydroxide ions, combine them then find the excess hydroxide, from there you can find the pH, use the ICE method.

The relevant equation is

$$H^+ + OH^- → H_2O$$

3. Apr 20, 2014

### Staff: Mentor

My understanding is that ammonium hydroxide is a weak base. The Kb is pretty low, and NH4OH appears on several weak base lists that I've found.

Chet

4. Apr 20, 2014

### Ritzycat

Yes ammonium hydroxide is a weak base. I don't think we've tackled strong acid-base problems yet. This is what I have done so far.

NH3 + H2o <-> NH4+ + OH-

[NH3] = (.006-.004)/.04 = .05

This is based off of the number of moles of NH4OH minus the number of moles of HCL divided by the total volume of the reaction.

Have I done it right so far?

5. Apr 20, 2014

### Staff: Mentor

If the HCl had reacted completely with the NH4OH, you would have remaining in the solution, NH4OH, and NH4+. So assume you have these to start with, and then the system has to re-equilibrate. The basic reactions are then:

NH4+ <->NH3 + H+

NH4OH<->NH4+ + OH-

H+ + OH- <-> H20

Other reaction equations you write are sums of these.

Chet

6. Apr 21, 2014

### Staff: Mentor

7. Apr 22, 2014

### Ritzycat

I used the first two formulas you gave me and using the concentrations of the NH4 and NH4OH I ICED the two and got

[H+] = 1.3x10^-5
[OH-] = 2.3x10^-3

I know that I can simply find the pH now, but how do I "combine" these to get the balanced pH of the entire reaction, not just the two smaller reactions?

I have a feeling its something to do with that h2o equation.... and 1x10^-14...

Last edited: Apr 22, 2014
8. Apr 22, 2014

### Staff: Mentor

You can't do the two reactions individually. You have to do the problem with the three reactions combined. To do this, you need to use the equilibrium constants for the three reactions. What are the equilibrium constants for the first two reactions? If the HCl + NH4OH reaction had gone to completion, what would the concentrations of NH4+ and NH4OH have been? This would be your starting point for solving for the re-equilibration to get final steady state.

Chet

9. Apr 22, 2014

### Ritzycat

Ka of NH4+ = 5.6 X 10^-10

Kb of NH4OH = 1.78x10^-5

what do I do with these numebrs?

10. Apr 22, 2014

### Staff: Mentor

If the original reaction had gone to completion, the concentration of NH4OH would have been 0.05M, and the concentration of NH4+ would have been 0.1 M. Do you see where I got these numbers from? After this, we are going to let the reactions re-equilibrate, starting from these as initial concentrations.

Let the reaction NH4OH <-> NH4+ + OH- react forward x moles/liter to reach final equilibrium
Let the reaction NH4+ <-> NH3 + H+ react forward y moles/liter to reach final equilibrium
Let the reaction H+ + OH- <-> H2O react forward z moles/liter to reach final equilibrium

So, the final concentration of NH4OH would be 0.05 - x
The final concentration of NH4+ would be 0.1 +x -y
The final concentration of NH3 would be y
The final concentration of OH- would be 10-7+x - z
The final concentration of H+ would be 10-7+y-z

These concentrations must satisfy the three equilibrium equations. What approximations do you think you can make to simplify the solution?

Chet

11. Apr 22, 2014

### Ritzycat

Thank you for your detailed response. I think I understand how you got those initial concentrations, but for the second part, did you simply ICE each of those equations to get those final concentrations, or what did you do? I've never used several variables before so I don't really know what to do with these.

12. Apr 23, 2014

### Staff: Mentor

People, please, why so complicated approach? It is apparently way above Ritzycat head. It is a classic one minute plug and chug exercise solved by high school students, you are nonsensically overcomplicating it!

Just assume ammonia protonation went to completion, that means you can easily calculate concentration of ammonia left and ammonium ion present (hint: what is a limiting reagent here?). The latter is a weak acid, the former is the conjugate base, so you have a buffer. Plug these concentrations into the Henderson-Hasselbalch equation and you have an instant answer.

I just checked - I got the answer in less than a minute. Seriously.

13. Apr 23, 2014

### Ritzycat

No need to be rude. I already am not feeling well about any of this, it is not helping for you to harass me about it, and that is exactly why I am afraid to say I do not understand something on this forum, as with my teacher, who would do the same. You do not have to "help" me.

I used the equation thing for the base, and I got 2.6 for the pOH so the pH is 11.4.

14. Apr 23, 2014

### Saitama

Hi Ritzycat!

That equation would have been useful if it was an acidic buffer. But we are dealing with a basic buffer here. Have a look at the link Borek posted in his previous post. :)

If you do not prefer to go by plug and chug, start with the definition of $K_b$. In the given case,
$$K_b=\frac{[NH_4^+][OH^-]}{[NH_4OH]}$$
You know $K_b$, $[NH_4^+]$ and $[NH_4OH]$. Can you find $[OH^-]$? ;)

EDIT: Ah, I see it now that you have reached an answer but that doesn't look right to me.

Last edited: Apr 23, 2014
15. Apr 23, 2014

### Ritzycat

I viewed that page but we have not even learned yet the material that these equation(s) are utilising.

in that equation, [OH-] would be .0023
and -log10(.0023) is 2.64 and that makes the pH 11.36.

16. Apr 23, 2014

### Saitama

I don't get that value for [OH-], it helps if you show the working.

17. Apr 23, 2014

### Ritzycat

Kb = 1.78x10^-5
[NH4+] = 2.3x10^-3
[NH4OH] = .3

(1.78x10^-5) x (.3)

then

(5.3x10^-6) / (2.3x10^-3)

That gives me .0023.
I got the value of NH4, which i feel may be wrong here, by icing:
NH4OH <-> NH4+ + OH-

18. Apr 23, 2014

### Saitama

The concentrations are incorrect. :/

I get $[NH_4^+]=0.1\,M$ and $[NH_4OH]=0.05\,M$. What are the moles of NH4+ and NH4OH after the addition of HCl? What is the final volume?

19. Apr 23, 2014

### Staff: Mentor

I don't think there was anything rude or harassing in my post, sorry if you read it this way. What I was aiming at was the fact Chet was guiding you into the deep forest instead of showing wide and well used highway.

His approach should in the end yield the same answer, but it doesn't make sense to go there, when there is a much easier route.

20. Apr 23, 2014

### Staff: Mentor

Hi guys.

Borek. Thanks for your comments, but I haven't worked a problem like this for almost 50 years, so I wasn't comfortable making the approximations that you are totally comfortable with; and I'm not familiar with Henderson-Hasselbach at all.

Panov-Arora seems to have gotten things back on track. Relative to my previous post, the approximations implied in his development are x << 0.05, and (x-y)<<0.1. This leads immediately to the value of [OH-]. Then the value of [H+] follows directly. Once these are known, one can calculate [NH3], and then confirm that the above approximations were valid. In my previous posting, I was trying to get Ritzycat to recognize the possibility of these kinds of approximations, and then use them to get the final answer. But, Borek, you were right; doing the problem my way was probably a little beyond his current mathematical level. I apologize to Ritzycat for getting things this complicated. However, in my defense, it was the only way I could be sure myself of getting the right answer.

Chet