Let me explain how I re-did this problem. What I posted originally was incorrect. Now, I'm not sure what I'm going to post is entirely correct either, so I'd really appreciate an extra set of eyes here looking over what I'm about to do below :)!
1) pH indicators work through acid-base chemistry. Let's say HX is a hypothetical indicator. HX happens to be an acidic indicator.
HX (white) + H_{2}O \leftrightharpoons X^{-}(black) + H_{3}O^{+}
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, K_{a}, equals \frac{[H_{3}O^{+}][X^{-}]}{[HX]}.
We can easily calculate K_{a} understanding that the at pH of 8.0:
[H_{3}O^{+}] = [X^{-}] = x = 1 × 10^{-8}
and
[HX] = M_{i} - x
so
K_{a} = \frac{[H_{3}O^{+}][X^{-}]}{[HX]} = \frac{x^{2}}{M_{i} - x}
and because we're at the halfway point, M_{i} - x = x, so
K_{a} = \frac{[H_{3}O^{+}][X^{-}]}{[HX]} = \frac{x^{2}}{x} = x
thus,
K_{a} = 1 × 10^{-8}
Now, in the 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:
K_{a} (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 K_{a} (HA) to be able to compare the strengths of the two acids. We can use the framework I outlined above:
K_{a} (HA) = \frac{[H_{3}O^{+}][A^{-}]}{[HA]} = \frac{x^{2}}{M_{i} - x} = \frac{x^{2}}{0.004 - x}
Since
pH=−log[H_{3}O^{+}]
it follows that
10^{-pH} = [H_{3}O^{+}] = 10^{-2.5} = x
and thus:
K_{a} (HA) = \frac{(10^{-2.5})^{2}}{0.004 - (10^{-2.5})} = 1 × 10^{-2}
In conclusion the strengths of the two acids are the same because we arrive at identical K_{a} 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 K_{a} (HA) value to one significant figure, and both acid ionization constants are identical to one significant figure.
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That was a long, drawn out, and painful process. Did I do everything right? I checked my work a few times and I don't think I made any faulty assumptions or any math errors. I think I sussed out every error I made on my third time around doing this problem (I've been tackling this problem for the last few hours!) Still, I'm not infallible, and I'd love to hear some advice or tips or suggestions or anything regarding my method and its accuracy!
