Kaplan Titration Question!

[RHC1987]

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)

Answer:
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!)

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[BloodySurgeon]

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)

 

[Kaustikos]

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.

 

[eeyoreDO]

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

 

[BloodySurgeon]

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

he meant 10^-2

 

[eeyoreDO]

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, indicating the presence of excess protons for methyl red to bind to.

Our pH indicator shows at around 5 though, not 7

 

[BerkReviewTeach]

Our pH indicator shows at around 5 though, not 7

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

 

[Kaustikos]

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

Thanks.
And I edited my post.

 

[RHC1987]

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!

 

[Kaustikos]

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!

The equation is great for determining the ratio of acid to base for whatever compound of interest (insert compound pka you have). The equation basically lets you determine (from what is known/unknown) the pH of a solution from a known base/acid and pka, ratio of acid to base from pka and pH or even the pka of that compound from setting a certain pH and looking at the ratio of acid to base. The way I like to look at it - the pH is the environment/variable in the equation (in this case) and the pKa is a constant. How will the pH affect the concentration of conjugate base/acid? You would need to use the pKa of methyl red because it tells you at what "pH" there would have to be in order to have a 1:1 ratio of conjugate base to acid. It's something "known" about the compound of interest.
I guess I do not understand why you don't see how you cannot use the pka of methyl red?
The pka is used because we're trying to determine whether or not methyl red is protonated or unprotonated in comparison to its environment. The methyl red will be largely protonated at pH of under 5.

For this question, though, you would not have to use that equation to know what would happen. It's great for clarifying things if you're really hazy on what the real answer is, but conceptually, one can determine what goes on without it.