[norwegiandude]

Hello everyone! I'm new to this forum, so just tell me if I post at the wrong place with the wrong subject or whatever, but without more delay, the question!

I have 200 L 0,1 M CH₃COOH, and then I add 5 grams KOH.

The pH of 0,1 M CH₃COOH was 2,875... ish from my calculations.
(I know pH is supposed to only have 1 decimal, but I dragged in more to be able to show the difference later on)

I assumed 1 mol OH^- neutralized 1 mol H₃O⁺

So I get a new concentration of CH₃COOH, at 19,911 M after adding 5 grams (0,089 mol) KOH ??

Anyways, I then calculated new pH to be 2,876  , a tiny difference of 0,001 pH.


Now, have I done anything wrong in the calculations, or is this the correct answer?


Sincerely,
norwegiandude

[Borek]

You can't ignore presence of the neutralized anion.

Technically it is a buffer solution, see examples of calculation here: http://www.chembuddy.com/?left=buffers&right=composition-calculation

In short: you should calculate concentrations of acetic acid left and acetate produced, and plug them into Henderson-Hasselbalch equation.

Or, you can calculate concentrations of acetic acid left and acetate produced and use ICE table to find the concentration of H⁺ in such solution. The latter method is marginally more accurate.

Are you sure it is 200 L, and not 200 mL? That would make it a limiting reagent problem.

[norwegiandude]

I'm positive its 200L, and now i used the Henderson-hasselbalch, getting a pH of 4,743. Just above the pKa for acetic acid.
 
Though I'm not so sure i used the right approach here.

The reason I'm so unsure, is because of the amount of acid here. Will the 5g KOH really have a noticable change in pH?

[Borek]

I'm positive its 200L, and now i used the Henderson-hasselbalch, getting a pH of 4,743. Just above the pKa for acetic acid.

That's for sure wrong, show what you did.

The reason I'm so unsure, is because of the amount of acid here. Will the 5g KOH really have a noticable change in pH?

Define "noticeable".

Chances are I was initially wrong and HH approach can be misleading, perhaps ICE table should give the better result. Still, please show how you got your pH value.

[norwegiandude]

I'll show the calculations when I get backhome, thanks for the help so far!

[Babcock_Hall]

"19,911 M"  Surely this is a typo of some kind.

[Borek]

He meant 19.911 moles of the acid.

[norwegiandude]

"19,911 M"  Surely this is a typo of some kind.

Yes, it was a typo, I meant 19,911 moles.

That's for sure wrong, show what you did.

I used the Henderson-Hasselbalch, but I see what i did wrong at the last effort. I put the value of OH^- moles at 20,089 for some reason, which obviously is wrong.

                                       ( 20,089/200 )
Getting the: pH = 4,74 + log (------------) = 4,743
                                       ( 19,911/200 )

Now, I did try it again, but this time, I did not do the same mistake, and only had 0,089 moles instead of 20,089 moles. But the answer i got then, was a pH all the way down on 2,4?? And that would just be weird.

Define "noticeable".

With noticable, I meant like something around my first answer, where you could see the pH change atleast a little (0,001 change), in comparison to, for example, a change of 0,0000000001 in pH.

[Borek]

Now, I did try it again, but this time, I did not do the same mistake, and only had 0,089 moles instead of 20,089 moles. But the answer i got then, was a pH all the way down on 2,4?? And that would just be weird.

Yes, and that's a clear sign that this approach is unfortunately not working (as it seems to suggest pH goes down from the original value after adding KOH).

That means we need another approach. I would go for an ICE table:

http://www.chembuddy.com/?left=buffers&right=with-ICE-table

Note: this is exactly the same problem you are trying to solve, as adding some KOH to acetic acid is equivalent to mixing acetic acid with potassium acetate.

With noticable, I meant like something around my first answer, where you could see the pH change atleast a little (0,001 change), in comparison to, for example, a change of 0,0000000001 in pH.

My estimates say difference around 0.1 pH unit, so it should be noticeable.

[norwegiandude]

For some reason, I find the ICE table really confusing in this case, maybe because I don't speak english that much or the descriptions are hard to understand for me.

My formula came out like this:
                                  x (20,089 + x)
1,8 * 10^-5 =  -------------
                                      19,911 - x

Giving me : x² + 20,089018x - 3,6*10^-3

Where I get: x= 1,79*10^-5   ------>  pH = 4,747


Am I using wrong numbers here? I'm feeling this problem shouldn't really be this hard, haha! Thanks for still helping though!

P.S.: Tried using just 0,089 instead of 20,089, and got pH = 2,41 (Thought you should know)

[Borek]

1. You forgot to divide by 200 L.

2. Initial amount of acetate is 0.089 moles (that much acetic acid was neutralized by KOH), not 20.089 moles.

[norwegiandude]

Ok, did it, and got pH 2,948, a change of ~0,075...! Thank you so much for helping! And just to make it clear, I actually skimmed the surface of that equation earlier in my rambling thoughts around the problem, but sometimes, it just gets too messy on both paper and in my mind to get any results.

Also happy about you not just giving me the answer straight off, now i actually learned something. Thumbs up!

So, do i check this post as "resolved" or something now?


Now that I'm done with this, I can finally continue reading about Fungi, which I haven't been able to concentrate about due to this problem.

[magician4]

Ok, did it, and got pH 2,948, a change of ~0,075...!

this to me seems to be wrong
(either which way the  calculation method your'e using: this is way off)

regards

Ingo

[norwegiandude]

The pH of 200 litres 0,1 M acetic acid is 2,875. And after adding 5g KOH, I calculated (with help from Mr. pH) new pH to be 2,948.

Can you support your claims?

[magician4]

yes, I can

5 g KOH equals 0.0891 moles of KOH = 8.91 * 10^-2 moles
diluted in 200 L this equals c =  4,455 * 10^-4 moles/L

this leaves 0.1 moles/L - 4.455 * 10^-4 moles/L = 9.955 * 10^-2 moles/L  of acetic acid

two ways to calculate for this, becoming more accurate thereby:

(a) c(CH₃COOH) >> c(CH₃COO^-) , therefore, reprotonation of acetate can safely be neglected, and the situation can be calculated according to "pure" acetic acid, 9.955 * 10^-2 moles/L , with simplified formula:

[tex]pH = 0.5 * ( pKa - log c ) = 0.5 * ( 4.75 - log 9.955 * 10^{-2} ) = 2.87597... [/tex]
this, obviously, is not a very good result  we'd better take the reprotonation into account

(b) ... and we're doing this by using the full LMA expression:

[tex]K_a = \frac { ( 4.455 * 10^{-4} + [H^+]) *[H^+]}{9.955 * 10^{-2} - [H^+]}[/tex]
[itex]\to [H^+]  = 0.00111889...     ,  pH = 2,95121... > 2,948 [/itex]  it differs, beyond any error of rounding


regards

Ingo