The indicator methyl red has a pKHIn = 4.95. It changes from red to yellow over the pH range from 4.4 to 6.2.
If the indicator is placed in a colourless buffer solution of pH = 4.55, what percent of the indicator will be present in the acid form, HIn?
What colour would you expect the solution in question 13 above to most look like?
Select one:
a. red
b. colourless
c. magenta
d. yellow
pH = pKa + log (In^-/HIn)
4.55 = 4.95 + log (In^-/HIn)
-0.4 = log (In^-/HIn)
(In^-/HIn) = 0.398 or
(In^-) = 0.398(HIn) so when (HIn) = 1, (In*-) = 0.398
%HIn = [(HIn)/(HIn) + (In^-)]*100 = [1/(1+0.398)]*100 = ?
What color is it at this pH? I don't know because I've never done a titration like that however, it is largely red. So red (a) is mostly right. It will not be b or d. You would expect it to be red to yellow or orange at about 5 or 5.1 So the almost all red will have some yellow or yellow orange mixed in with it. I don't think magenta has orange in it but I don't know. If I had to answer I would choose a as the answer since I don't have a clue what magenta is. :-)
Disregard this answer.
I believe this is an easier way to explain.
pH = pKa + log [(In^-)/(HIn)]
4.55 = 4.95 + log[(In^-)/(HIn)]
-0.4 = log [(In^-)/(HIn)]
In^-/HIn = 0.398 and
In^- = 0.398*HIn is equation 1
equation 2 is In^- + HIn = 1.0
solve the two simultaneously to obtain
0.398 HIn + HIn = 1
HIn = 1/1.398 = 0.715 and
In^- = 1 - 0.715 = 0.285
So HIn percent = 71.5% and In^- = 28.5%
For the color:
When pH = 4.95 we have base/acid = 1 so acid and base are @ 50%.
At the pKa value of 4.95 we have a 50-50 mixture of red and yellow which makes orange. I am told that mixtures of red and yellow will not give magenta so I'm going with a red color @ pH 4.55 which is about 70% red and about 30% yellow
To determine the percent of the indicator that will be present in the acid form (HIn) in a buffer solution of pH = 4.55, we can make use of the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, pH is the given pH of the buffer solution (4.55), pKa is the negative log of the acid dissociation constant (4.95 for methyl red), [A-] is the concentration of the indicator in the ionized form, and [HA] is the concentration of the indicator in the acid form.
We can rearrange the equation to solve for the ratio of [A-] to [HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values:
[A-]/[HA] = 10^(4.55 - 4.95)
[A-]/[HA] = 10^(-0.4)
[A-]/[HA] = 0.398
Now, we can calculate the percent of the indicator present in the acid form:
Percent HIn = ([HA]/[HA] + [A-]) * 100
Percent HIn = ([HA]/([HA] + 0.398[HA])) * 100
Percent HIn = ([HA]/1.398[HA]) * 100
Percent HIn = 1/1.398 * 100
Percent HIn = 71.53%
Therefore, approximately 71.53% of the indicator will be present in the acid form (HIn).
As for the color of the solution, the indicator methyl red changes from red to yellow over the pH range from 4.4 to 6.2. Since the given buffer solution has a pH of 4.55, which is within the range where methyl red appears red, we would expect the solution in question 13 to most look like the color red (option a).