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

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#### FutureScaresMe

##### UCSF '14
10+ Year Member

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

##### Full Member
Moderator Emeritus
10+ Year Member
5+ Year Member
Moving to study Q&A subforum

#### FutureScaresMe

##### UCSF '14
10+ Year Member
whoops 10+ Year Member
Bueller?

#### rnicholsaz

##### Full Member
10+ Year Member
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

##### Full Member
10+ Year Member
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

##### UCSF '14
10+ Year Member
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 Last edited:

#### FutureScaresMe

##### UCSF '14
10+ Year Member
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

##### New Member
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

##### Full Member
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]

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