# Chemistry pH Review

#### Teleologist

Self-studying chemistry here and ran across this question.

Indicator HIN is yellow at pH 1.0 or less, green at pH 2.0, and blue at pH 3.0 or above.

A 0.004 M sample of colorless acid HA is treated. It develops a green-blue color halfway between pure green and pure blue. How do the strengths of HA and HIN compare?

A) HA is weaker
B) HA is stronger
C) HA and HIN are of the same strength
D) Not enough information

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This is my thought process. The part about HX is part of my thought process and an introduction to how I approached this problem.

HX only works on a pH range of 7.0 to 9.0. The HX acid molecule is a snow white color, and its anion (conjugate base) is a midnight black.

At a pH of 7.0, there is more hydronium than acid initially, and by extension, there are also more acid anions. This causes the equilibrium to shift to the left, and the solution is white.

At a pH of 9.0 there is less hydronium, and by extension, fewer acid anions initially. The equilibrium position therefore moves to the right. So the solution is black.

At a pH of 8.0, there are equal concentrations of acid molecule and acid anion, so the solution is gray.

The acid ionization constant, Ka, equals [H3O+][X−][HX].

We can easily calculate Ka understanding that the at pH of 8.0:

[H3O+]=[X−]=x=1×10^−8

and

[HX]=Mi−x

so

Ka=[H3O+][X−][HX]=x^2/(Mi−x)

and because we're at the halfway point, Mi−x=x, so

Ka=[H3O+][X−][HX]=x^2/x=x

thus,

Ka=1×10^−8

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Now, in the ACTUAL problem, we're dealing with different indicators and acids, but the same principles apply. We have the indicator HIN in the problem, which works on on a range of pHs from 1.0 to 3.0. At 2.0 the the indicator is green - a equal mix of yellow and blue - which implies that at pH 2.0 there are equal concentrations of the acid and the acid anion (the acid's conjugate base). And so:

Ka(HIN)=1×10^−2

We now know the strength of the acid HIN.

We must now proceed to find the strength of the acid HA. We know that 0.004 M of the acid HA was treated with a drop of HIN and the pH of the sample of HA was determined to be 2.5 through colorimetric analysis.

Given pH we must now determine Ka(HA) to be able to compare the strengths of the two acids. We can use the framework I outlined above:

Ka(HA)=[H3O+][A−][HA]=x^2/(Mi−x)=x^2/0.004−x

Since

pH=−log[H3O+]

it follows that

10−pH=[H3O+]=10−2.5=x

and thus:

Ka(HA)=(10−2.5)20.004−(10−2.5)=1×10^−2

In conclusion the strengths of the two acids are the same because we arrive at identical Ka values. We are limited to one significant figure because in the problem we are given a 0.004 M HA solution. This number only has 1 significant figure since leading zeros don't count. Therefore we can only report the Ka(HA) value to one significant figure, and both acid ionization constants are identical to one significant figure.

#### Next Step Tutor

##### MCAT Guru
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5+ Year Member
What was the answer provided by the book?

I would've looked at that and probably just guessed (A) but marked it and moved on. (with a very quick, sloppy conceptual approach that at blue-green the indicator is mostly deprotonated, so the protons left the indicator and went onto the other molecule, thus if the indicator is the one that deprotonates, it's a stronger acid).

(Edited b/c I'm a sloppy ******* who didn't take the time to actually solve the problem. Ignore this post and read Berkeley's response below. He's got it right)

I tried following your explanation but there's something wrong with your formatting and it's not making sense to me.

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7+ Year Member

#### Teleologist

Jayoh,
I just converted the 0.004M to pH and saw that it was ~2.4, which was right in between the green and blue in the example and reasoned that they are the same strength. Maybe I drastically oversimplified?
I'm not clear what I did; did you convert molarity of HA to pH? How did you do that?

#### jayoh

##### turning mountains into molehills
7+ Year Member
Jayoh,

I'm not clear what I did; did you convert molarity of HA to pH? How did you do that?
Yeah, right after I wrote that I realized that I did simplify by assuming that the acid completely dissociates, but maybe that's a fair assumption for the mcat? So if the acid completely dissociates then it would be pH=-log(0.004) = 2.397

#### Teleologist

Yeah, right after I wrote that I realized that I did simplify by assuming that the acid completely dissociates, but maybe that's a fair assumption for the mcat? So if the acid completely dissociates then it would be pH=-log(0.004) = 2.397
Well, the pH of acid HA is 2.5, from coloriometric analysis. The problem says it develops a color halfway between the color corresponding to a pH of 2 and a pH of 3.

#### jayoh

##### turning mountains into molehills
7+ Year Member
Well, the pH of acid HA is 2.5, from coloriometric analysis. The problem says it develops a color halfway between the color corresponding to a pH of 2 and a pH of 3.

It's a hard problem, I don't know if my reasoning is right. But I guess my logic is that according to the question, if the strength of HA is about the same as HIn, then it would be about ~2.5 based on the color. It also tells you the molarity, which would translate to a pH of ~2.5. So they should be about the same strength.

#### Teleologist

It's a hard problem, I don't know if my reasoning is right. But I guess my logic is that according to the question, if the strength of HA is about the same as HIn, then it would be about ~2.5 based on the color. It also tells you the molarity, which would translate to a pH of ~2.5. So they should be about the same strength.
Oh yes, it's a hard problem. But yes I calculated that the Ka values for both acids were the same to one significant figure so therefore they were of the same strength.

How did you determine the strength of the indicator HIN?

#### getsome111

Im going to preface this with a little background information you guys might have failed to consider;

First thing that caught my eye was that its usually pretty unlikely to have 2 ionizable protons, in fact I've never even herd of an indicator with multiple ionizable protons, and especially not all around the same pKa. See where I'm going with this?

The color goes from Yellow @1, [email protected], [email protected] if you weren't a little intrigued by those close pKas then this would surely be your next hint.

The final piece of evidence comes from a background knowledge of a few key numbers I memorized awhile ago from the henderson hasselbach eqn(or whatever) that when pH=pKa the species is 50% protonated, when you are 1 point away from the pH it is either 9% or 91% bound depending on which direction you go in.

This all makes it a pretty clear case that the pKa of the indicator is @ph=2, its 2 forms are yellow and blue.

-but don't stop here, if you compare right now you'll be wrongfully led into thinking that the indicator is the stronger acid, I almost made this mistake

for the .004M species HA, which has an ability to create a Ph=2.4 solution IF it completely dissociates, keep in mind that you will only have 50% dissociation when pH=pKa, The indicator is telling us that the pH of the solution is around 2.5 so the HA must have a pKa AT LEAST 1 pH unit below 2.5 (that would give 91% dissociation or in other words 91% of those protons from the acid would be free to bring the pH of the solution down).

So I think I've figured it out, Now just compare pKas to find out which ones the stronger acid
pKa HA <1.5 (thus the stronger acid)
pKa Indicator 2

goodluck!

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#### BerkReviewTeach

##### Company Rep & Bad Singer
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10+ Year Member
I have to chime in. When it's a question from another company, I try to shy away out of etiquette. But there is some over-complication flying around and it's potentially harmful.

First: There are not two ionizable protons on HIn. What you are given in the question is that it's yellow at pH = 1 and blue at pH = 3, so its protonated form is yellow in color and deprotonated form is blue in color. When it's green, that means you have equal parts yellow and blue, so you are at a point where pH of the solution = pKa of the indicator. So the indicator has a pKa = 2.0, making it a weak acid.

Second: If HA were strong, it would fully dissociate in water, giving .004 M H+. -log (0.004) = 3 - log 4 = 2.4. The passage said it's half way between blue and green, which is saying that the pH = 2.5. We're working with an indicator to estimate the pH, so 2.4 and 2.5 are essentially the same. This means that HA is a strong acid, because it fully dissociated. Even if you want to get technical and consider the 0.10 difference, it still means that the HA was at least 90%-dissociated, which still makes it strong.

Given that the indicator (in its protonated form) is a weak acid and HA is a strong acid, choice B is your best answer.

#### Schenker

I think a lot of you guys are making this way harder than it needs to be. The indicator is protonated at a higher pH and deprotonated at a low pH. It is fully (99%) deprotonated at yellow and fully protonated at blue. The question says you have some blue character. So the indicator is more protonated than deprotonated. Which means it has a higher affinity for the free protons vs the acid = In- is a better base than A-, so HA is the stronger value. No lograthimic calculations needed The bolded is not correct. Your reasoning would have missed this question if ".004M sample of HA" were replaced with "1M sample of HA," as the answer in that case would be that HA is a weaker acid than HIn. The correct reasoning needs to address the molarity of the HA sample; any explanation simpler than BerkReviewTeach's explanation wouldn't be correct.