# Titration 1986 AP Exam

1. Feb 13, 2005

### r3dxP

This question appeared on 1986 AP Exam.

1) In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka, equal to 2.8E-5 at 25 degree C. A .300L sample of a .050molar solution of the acid is prepared.
Now.. there is a part d) to this question...
d) To the remaining .150 liter of the original soln, .075L of .100molar NaOH soln is added. Calculate the [OH-] for the resulting solution at 25 degree C.

If you have any clue on how to do this, any help would be greatly appreciated. Thanks.

2. Feb 13, 2005

### Gokul43201

Staff Emeritus
This isn't homework, is it ?

First, you can assume that HN3 is essentially a monoprotic acid.

1. From the definition of Ka, you can find [H+].

2. You can then assume that NaOH is essentially 100% dissociated, so you know the [OH-] from NaOH.

3. The excess is the number that you need.

3. Feb 13, 2005

### r3dxP

No this is not my homework. I was reviewing old AP exams for my own benefit and came across this question that I did not understand. Now i figured out how to do it. Thanks for the help gokul.

4. Feb 23, 2005

### Staff: Mentor

It is a buffer question (Henderson-Hasselbalch equation).

I am afraid your explanation is faulty. Adding NaOH you move dissociation equlibrium, so you can't just calculate $$[H^+]$$ from the Ka and substract $$[OH^-]$$ concentration:

In 0.05M hydrazoic acid $$[H^+]$$ = 1.05e-3M and pH = 2.98

Following your line of thinking (and taking dilution into account), after NaOH addition $$[OH^-]$$ = 0.068M

so the excess $$[OH^-]$$ is 0.067M, and pH = 12.83

while the solution after mixing is slighty acidic, with pH = 5.13. NaOH was used to neutralize 75% hydrazoic acid.

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis
All calculations in my posts are done using BATE

5. Feb 23, 2005

### Gokul43201

Staff Emeritus
I don't understand how it is, since we are at the equivalence point (0.15L*0.05M = 0.075L*0.1M) and all the acid would have been converted into (0.0333M) sodium azide. I did however, fail to consider the hydrolysis of the azide. Taking tihs into account, I get [OH-] ~ 0.35 E(-4) or pH ~ 9.54

Now, in all likelihood, I'm wrong again - it's been nearly a decade since I worked on these problems and I've since gotten rusty - since you clearly know what you're talking about, and I'm only going by what I think makes sense to me.

6. Feb 23, 2005

### GCT

Borek

What Gokul suggested was correct, this is not technically a buffer problem which frequently employs an near equal amount of its conjugate salt of the weak acid. Also henderson-hasselbach is not necessary in any case, you can simply solve this by employing the equation for Ka.

7. Feb 23, 2005

### Gokul43201

Staff Emeritus
Actually, I believe it would be a buffer (requiring the use of HH) for nearly any volume of NaOH up to the equivalence point, because until then you would have hydrazoic acid and sodium azide (conj. salt) in the solution.

8. Feb 23, 2005

### Staff: Mentor

My mistake - somehow I mixed volume with concentration and I did the calculations not for 0.1M NaOH but for 0.075M, thus I was before end point, just in the buffer range. Please remember I am in Europe, so when you are still fresh it is already a.m. here :zzz:

Still, it seems to me your answer is wrong - endpoint pH should be 8.58, not 9.54.

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis
All calculations in my posts are done using BATE

9. Feb 23, 2005

### GCT

A buffer relates to the common ion effect, thus if you were to mix an aqueous solution of the acid and a fair amount of an aqueous solution of its conjugate salt, what you'll have is a lower concentration of hydronium ions in the end. This is not a buffer problem. Also remember that the h-h equation is derived from pKa, it's just a more convenient form when it comes to making buffers.

10. Feb 24, 2005

### GCT

To explain at bit further,

$$Ka= \frac{ [N^{_3-} ] [H_3O^{_+}]}{[HN_3]}$$

What happens with buffer problems is that in addition to the initial concentration of each ion determined purely by the dissociation of the acid, we modify it so that we take into account the extra additive concentration of the conjugate base when we add the salt. Occasionally buffers are created by simply dissolving the conjugate salt in aqueous solution, however we have simply dissolved an acid in this case. It's a simple Ka problem.

Last edited: Feb 24, 2005