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Finding pka from pH

  • #1

Homework Statement


When 100.0 mL of weak acid HA was titrated with 0.09381M NaOH, 27.63mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What is the pKa of HA

Homework Equations


pH=pKa+log([A]/[HA]

[OH]-10^(-pOH)

The Attempt at a Solution



14-10.99=3.01 this gives me pOH

From this I can calculate the concentration of OH: 0.000977237 M

Since we are at the equivalence point, to get the moles I multiplied the above concentration by 0.02763L and got 2.7*10^(-5)

I know that initially I had 0.0025919703 moles of HA

I subtract the moles of OH that react with HA and get 0.0025919703-2.7*10^(-5)=2.56399*10^(-3)

The moles of OH are the moles of A: 2.7*10^(-5)

Using pH=pKa+log([A]/[HA]

I get 10.99=pKa-1.97

Giving me a pH of 12.96 instead of 9.69 which is the correct answer

What am I doing wrong
 

Answers and Replies

  • #2
Borek
Mentor
28,326
2,705
I subtract the moles of OH that react with HA and get 0.0025919703-2.7*10^(-5)=2.56399*10^(-3)
Doesn't make much sense. OH- can't be at the same time present in the solution (pOH) and be the one that reacted with the acid. Either, either.

The moles of OH are the moles of A: 2.7*10^(-5)
No.

Hint: pH at the equivalence point is that of a salt of a weak acid. Can you write reaction that is responsible for the OH- production in such a solution?
 
  • #3
mjc123
Science Advisor
Homework Helper
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424
to get the moles I multiplied the above concentration by 0.02763L and got 2.7*10^(-5)
Are you using the correct volume. Remember you have added the NaOH solution to the HA solution.
But Borek's point still applies.
 
  • #4
epenguin
Homework Helper
Gold Member
3,623
697
Can I ask what country you are studying in? This question has certain geographical hallmarks.I would probably fail the exam there because I would write as answer " good luck with finding the pH at the equivalence point!".I would know where the equivalence point is alright but the pH there is varying wildly. Anyway you have worked out the molarity of the NaA solution, now suppose we dissolved NaA In water to make that molarity, to ask what the pH of that solution would be, knowing that pKa, would make sense. So would the inverse question, knowing the pH what is the pKa?

It is too late at night for me to concentrate on your calculation, but there might be a mistake where you say the moles of OH-, are the moles of A?? - rather they are the moles of moles of HA aren't they? I suggest you write out the formula for Ka explicitly and then play with that – you should get a formula with a square in it that you may remember.
 

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