# Kaplan Titration Question!

Discussion in 'MCAT Study Question Q&A' started by RHC1987, May 11, 2008.

1. ### RHC1987

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Methyl red (pKA = 5.0) is a pH indicator dye that is red when protonated and yellow when deprotonated. A student chose to use methyl red as an indicator in her titration of an unknown concentration of HCl into a 250mL solution of 10-2 M NaOH. What will be the color of the solution before the titration begins, at the equivalence point of the titration and at the end point of the titration? (Assume the temperature of the solution remains constant at 25°C)

The solution will be yellow before the titration, yellow at the equivalence point and red at the end point.

(I'm having a hard time understanding whats going on at the equivalence point of the titration... Any feedback will be awesome!)

3. ### BloodySurgeon Moderator EmeritusLifetime DonorClassifieds Approved

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I am going to say the short version:

pOH= -log [OH]
pOH=2
pH + pOH = 14
pH = 12

pKa = 5.0
start: pH = 12 .......... deprotonated (yellow)
Equivalence pt: pH about 7 b/c OH- + H+ --> H2O ..... deprotonated (yellow)
end pt: pH 2 b/c nearly all HCl ................ protonated (red)

pH>pKa : deprotonated (yellow)
pH=pKa : half protonated, half deprotonated (orange)
pH<pKa : protonated (red)

4. ### Kaustikos Archerize It

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What the above poster explained is correct, but I thought I would try and help with your equivalence question.

At equivalence, you have protons but they're still being bound to the OH from the sodium hydroxide to form H2O. The only thing is that at equivalence, you have a pH of 7 due to the formation of H2O. Thus, you will still have a yellow color. Once you start exceeding the equivalence point is when you will have the red color showing in your titration - because of the methyl red protonating.

5. ### eeyoreDO

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What does the 10-2 in 10-2 M NaOH mean?

6. ### BloodySurgeon Moderator EmeritusLifetime DonorClassifieds Approved

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he meant 10^-2

7. ### eeyoreDO

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Our pH indicator shows at around 5 though, not 7

8. ### BerkReviewTeach Company Rep & Bad Singer Exhibitor

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That's why it was a poorly chosen indicator, which is perhaps a point being made in the question.

9. ### Kaustikos Archerize It

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Thanks.
And I edited my post.

10. ### RHC1987

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Yeah, that makes a lot of sense. I guess I got confused because the explanation offered to me dealt with the Henderson Hasselbach equation and I didnt know why the pKa of Methyl Red was used.
Here is the explanation they provided me:

The Henderson-Hasselbach equation may be used to determine the ratio between the acid and conjugate base of the indicator:

pH = pKA + log ([conjugate base]/[acid])
7 = 5 + log ([conjugate base]/[acid])
2 = log ([conjugate base]/[acid])

The pH at the equivalence point of a strong acid-strong base titration will always occur at about 7.00. Plugging into the Henderson-Hasselbach equation you will find that the ratio of conjugate base/acid will be 100:1; therefore, the solution will remain yellow at the equivalent point. Note that the student chose a poor indicator, for the purpose of an indicator is to begin changing color at the equivalence point. For a strong acid-strong base titration an indicator with a pKA of around 7 would have been a better choice. By definition, the end point of a titration is the pH at which the indicator remains permanently at its end color; therefore, its color must be red in this case. Specifically, this will typically occur at a pH one unit below the pKA as the ratio of conjugate base to acid will be 1:10 at that point.

I'm having a hard time understanding the equation, and why you can use the pKa of Methyl Red with the pH of the solution... Theres a very good chance I'm just confused with the terms. But thank you for the help so far!

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