Calculating pKa when pH is known

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Heres a question involving what I'm talking about

When 5mls of 0.05M CH3COOH is allowed to react with 1.5mls of 0.05M KOH. The pH is found to be 4.43. Using the Henderson-Hasselback equation calculate the pKa.

My teacher told me that you can find the amount of dissociated ions of the weak acid without even knowing the dissociation constant by writing out the reaction equation for the reaction with the base and if its a 1:1 reaction for example the amount of dissociated ions will be equal to the amount of the base that was added. Why is this? Why would the amount of dissociated acid ions be equal to the amount of base I add?
 

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  • #2
epenguin
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We won't answer till you show us you have done what the teacher told you to do! :smile: If you do you may see the answer yourself.
 
  • #3
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Heres my calculation
First I calculate the moles of CH3COOH in 5ml of 0.05M solution:
(0.05 / 1000) * 5 = 2.5 x 10-4 moles

Then the moles of KOH in 1.5ml of 0.05M solution:
(0.05 / 1000) * 1.5 = 7.5 x 10-5 moles

The balanced reaction equation is
CH3COOH + KOH -> KCH3COO + H2O

Since the stoichiometric ratio is 1:1 I know that the moles of dissociated acetate ions must be 7.5 x 10-5.

To get the amount of acetic acid remaining I subtract the moles of KOH from the initial amount of acetic acid
2.5x10-4 - 7.5x10-5 = 1.75x10-4

Now I use the Henderson-Hasselbalch equation pKa = pH + log([A-]/[HA]) to calculate the pKa.

pKa = 4.43 + log(0.000025/0.000175)
= 4.57

Ka = antilog(4.57)
= 3.72 x 10^3
The real pKa of acetic acid is around 4.7 so I wasn't too far off. I still don't know why the acetate ions are equal to the KOH though. If I react some acetic acid with KOH is it only the dissociated acetate ions that are involved in the reaction?
 
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  • #4
epenguin
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OK, you know, I hope, that KOH is a strong base. There is really no KOH molecule in aqueous solution, all the potassium there always exists as K+ only.

So by now you should know the concentration of all the charged species you have in that solution. Please list the electrically charged species present, and their concentrations as far as you can from what you have got already, then we'll come back on it.

Slight mistakes here in Ka = antilog(4.57) = 3.72 x 10^3 I think, you need to straighten. It's antilog(- Ka) plus you'll run into confusions if you don't get the habit of quoting physical parameters with their units, in this case molarity.
 
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All I know is that acetic acid is a weak acid and it doesn't dissociate fully into acetate and H+ ions when dissolved. I know KOH is a strong acid. All the charged species are K+ and H+ cations as well as OH- and CH3COO- anions. I know exactly how many K+ ions are present and ignoring the dissociation of water know exactly how many OH- ions present but what I don't know is how many acetate ions are present because acetic acid dissociates partially depending on its Ka.

If I react say 2 moles of KOH with 5 moles of CH3COOH then I know some potassium acetate will be formed and if I had to measure the ion concentration myself I'd do a titration and determine the concentration of KCH3COO because that would tell me the amount of acetic acid that DIDNT dissociate and then I could easily find out the percentage of it that did but this isn't the method we used in the lab.
 
  • #6
epenguin
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I know exactly how many K+ ions are present and ignoring the dissociation of water know exactly how many OH- ions present but what I don't know is how many acetate ions are present because acetic acid dissociates partially depending on its Ka.
Yes but if you
Please list the electrically charged species present OK you've done that bit right, and their concentrations as far as you can from what you have got already
I'm sure I can help. :smile:

I won't insist you list the equations you think come into this but you can if you like. :biggrin:

Edit: there are three: electroneutrality, euilibrium equation (aka mass action law, aka definition of equilibrium constant) and cosnervation of mass. Get at least the first which is key to your problem with this.

Also you do need to sort out/revise logarithms, definitions pH, pK.
 
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  • #7
DrDu
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What your teacher told you is an approximation (a very good one), but the problem can also be solved without making this approximation. Maybe you should try to figure it out yourself, it is quite instructive. Try to express the concentrations of Ac- and HAc in terms of the "initial" number of moles of HAc and KAc and resulting concentrations and of the concentration of H+. With the initial number of moles I mean the number of moles of HAc (dissociated and undissociated) and KAc you would have to mix to get the solution
 
  • #8
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I submitted that assignment already I'm not asking for help with that question I'm trying to learn the concept. If I add 15 tons of pure NaOH to a 10 litre bucket of 0.0005M CH3COOH solution theres no way the amount of acetate ions present is going to equal the god knows how many moles of NaOH in 15 tons.

Does the strong base force the weak acid to dissociate by any chance? If thats the case I know whats going on here otherwise its a complete mystery to me.
I suppose I better just learn about pH, pK and equilibrium reactions in more detail. I'd been meaning to do that for a good while anyway.
 
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  • #9
Borek
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Does the strong base force the weak acid to dissociate by any chance?
That's more or less correct. Unless acid is extremally weak, it is almost quantitatively neutralized by strong base.

For example if you prepare solution that is initially 1M in NaOH and 1M in acetic acid, only about 4e-5M of acetic acid will be left unneutralized - for most purposes that means 100% nautralization, although it also means that due to hydrolyzis pH is around 9.

Identical solution of much weaker acid - HCN, pKa=9.3 - will be neutralized in 99.3%.

--
 

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