Equivalence Point - Neutralization Point ? I'm alittle confused now

[FutureScaresMe]

So I was doing a practice exam, and came across a question which was asking about a diprotic acid with Pka1 of 6.37 and pka2 of 10.25. The question then asked what is true at 8.31. Well, I knew that the avg of pka1 and pka2 would be the equivalence point of the first acidic proton, which is 8.31. I remembered that the equivalence point is the point of neutrality, but that doesn't necessarily mean that the pH is 7 though.

So I chose the answer, the solution is pH neutral. The actual answer was [H30] = [OH]. But doesn't that also mean the pH is 7? Aghh, I am doing really well on PS, and this titration question is really irking me. Someone please point out the obvious here for me. Thank you!


Question 2:

If pKa for acetic acid is 4.7, then at pH of 5, shouldn't CB > Acetic Acid ?? The answer said acetic acid > CB

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[iA-MD2013]

Moving to study Q&A subforum

 

[FutureScaresMe]

whoops

 

[FutureScaresMe]

Bueller?

 

[rnicholsaz]

pH = pKa + log (A-/HA)

5 - 4.7 = log (A-/HA)

.3 = log (A-/HA)

10^1/3 = A-/HA

1/1000 = A-/HA

Hope that helps! I'm gonna have to go over acid base stuff tomorrow, that annoys me that I don't know the answer to the first one. Good luck on the test!

 

[rnicholsaz]

You said the pH was 8.31, then the solution cannot be pH neutral. At pH 8.31 you have neutralized all of the acid to the conjugate base, therefore the [-OH]=[H+]. If the concentration of acid and base are equal then you have reached the first equivalence point. If you had 2x the [-OH] than [H+] then you would reach the second equivalence point. Hope that helps.

 

[FutureScaresMe]

pH = pKa + log (A-/HA)

5 - 4.7 = log (A-/HA)

.3 = log (A-/HA)

10^1/3 = A-/HA

1/1000 = A-/HA

Hope that helps! I'm gonna have to go over acid base stuff tomorrow, that annoys me that I don't know the answer to the first one. Good luck on the test!

Yes that does help, thank you. But your calculation was wrong, confirming my original thoughts however. 10^(1/3) ~ 2, therefore the ratio of A^-:HA is approximately 2:1. So yes, base should be greater than acid concentration at a pH greater than the pKa.

Thanks for the help, didn't think to do it this way

 

[FutureScaresMe]

Yes, I realized after looking at the answer that 'pH neutral' implies a pH of 7. But the equivalence point can be considered as the point of neutrality, according to my review books. I was a little confused by the terminology. However, you said:

If the concentration of acid and base are equal then you have reached the first equivalence point.

Isn't this the definition of pKa when [HA] = [A-] ?

Again, I don't understand why the pH at equivalence of a titration between a weak acid and a strong base is greater than 7, if [OH-] = [H30+]. (Yes I know it's because the CB is formed, but I'm not satisfied with that answer).

 

[chaostheory44]

Yes, I realized after looking at the answer that 'pH neutral' implies a pH of 7. But the equivalence point can be considered as the point of neutrality, according to my review books. I was a little confused by the terminology. However, you said:


Isn't this the definition of pKa when [HA] = [A-] ?

Again, I don't understand why the pH at equivalence of a titration between a weak acid and a strong base is greater than 7, if [OH-] = [H30+]. (Yes I know it's because the CB is formed, but I'm not satisfied with that answer).

The definition of Pka is -logKa. The pH=pka when [HA]=[A-].

At the equivalence point of a weak acid titrated with a strong base, [OH-] does not equal [H30+]. Equivalence point occurs when all of the weak acid has been converted to its conjugate base. Because only the conjugate base remains, the pH will be greater than 7.

 

[Tarasaurusrex]

Question 2:

5 = 4.7 + log(A-/HA)

0.3 = log(A-/HA)

10^0.3 = (A-/HA)

2 = A-/HA

2[HA] = [A-]

Therefore the concentration of the base is 1/2 the concentration of the acid. This means the concentration of the acid is greater than the concentration of the base. This results in [A-] > [HA]