# TBR Titration (Equivalence Point)

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#### gopackers

##### Full Member
2+ Year Member
@BerkReviewTeach
So maybe I'm just having a brain-dead moment here, but TBR says we can estimate the equivalence point by averaging the pKa of the acid and the pH of the titrant. How do we know the pH of the titrant? In the example they used (Table 9.3 on Pg 100) the titrant was KOH which has a pH of 13. Is that something I should've had memorized or am I just missing something really obvious here? Thank you in advance and sorry if it's a dumb question.

#### SweetBurger

##### cowbell
10+ Year Member
I don't know why you need to estimate it. It's a pretty fast calculation to find the equivalence pH given you have concentration of the acid and the concentration of the KOH.

#### BerkReviewTeach

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15+ Year Member
@BerkReviewTeach
So maybe I'm just having a brain-dead moment here, but TBR says we can estimate the equivalence point by averaging the pKa of the acid and the pH of the titrant. How do we know the pH of the titrant? In the example they used (Table 9.3 on Pg 100) the titrant was KOH which has a pH of 13. Is that something I should've had memorized or am I just missing something really obvious here? Thank you in advance and sorry if it's a dumb question.

You calculate the pH of a basic solution by starting with pOH. The pOH = -log [OH-]. In the example you mentioned, the [OH-] is 0.10 M, so pOH = -log(0.10) = - (-1) = 1. If pOH is 1, then pH is 13 (from pH + pOH = 14). The equivalence point lies about half way between pKa and the pH once it's been well overtitrated (look at curve symmetry.) The pKa is 5, so the equivalence pH is 9.

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#### BerkReviewTeach

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I don't know why you need to estimate it. It's a pretty fast calculation to find the equivalence pH given you have concentration of the acid and the concentration of the KOH.

The calculation involves first determining the [conjugate base] at equivalence, and then using the ICE box to get the [OH-] concentration from that weak base solution. Then you calculate the pOH from there. Finally you need to subtract the pOH from 14 to get the pH. I will show the steps below.

(1) The titration involves titrating 0.10 M weak acid (with pKa = 5) with 0.10 M NaOH (aq). Because they are equal molarity, the volume of solution must be doubled once you reach the equivalence point. As such, we can quickly determine that the concentration of the conjugate base at equivalence is 0.050 M. We also know that it has a pKb of 9.

(2) Using the ICE box, we can find that [OH-] = square root (0.050 M x 10^-9) = square root (5 x 10^-11) = square root (50 x 10^-12) = 7.1 x 10^-6.

(3) pOH = -log (7.1 x 10^-6) = 6 - log 7.1 = 5.15 (this fast is using is a TBR shortcut that we recommend.)

(4) The pH = 14 - 5.15 = 8.85.

It is our opinion that averaging 5 and 13 to get 9 is faster and easier than the four-step method you suggest.

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#### SweetBurger

##### cowbell
10+ Year Member
The calculation involves first determining the [conjugate base] at equivalence, and then using the ICE box to get the [OH-] concentration from that weak base solution. Then you calculate the pOH from there. Finally you need to subtract the pOH from 14 to get the pH. I will show the steps below.

(1) The titration involves titrating 0.10 M weak acid (with pKa = 5) with 0.10 M NaOH (aq). Because they are equal molarity, the volume of solution must be doubled once you reach the equivalence point. As such, we can quickly determine that the concentration of the conjugate base at equivalence is 0.050 M. We also know that it has a pKb of 9.

(2) Using the ICE box, we can find that [OH-] = square root (0.050 M x 10^-9) = square root (5 x 10^-11) = square root (50 x 10^-12) = 7.1 x 10^-6.

(3) pOH = -log (7.1 x 10^-6) = 6 - log 7.1 = 5.15 (this fast is using is a TBR shortcut that we recommend.)

(4) The pH = 14 - 5.15 = 8.85.

It is our opinion that averaging 5 and 13 to get 9 is faster and easier than the four-step method you suggest.
Coming from an analytical chem background I don't perceive it to be particularly time consuming. But I won't disagree that averaging to get 9 is far faster and more efficient in terms of the MCAT. I'll be incorporating these techniques whenever possible to improve speed. Appreciate your reply.

#### gopackers

##### Full Member
2+ Year Member
Thank you both for your replies! I appreciate the help and I'm sorry for asking such a silly question

#### PlsLetMeIn21

##### Full Member
5+ Year Member
Their method for getting pH is so easy. I can say from my experience that it paid dividends on my MCAT. Know all of their shortcuts like this one!

10+ Year Member
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