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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
----
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
----
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.
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
----
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
----
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.
