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Relative Strengths of Acids

  1. Dec 14, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    [itex]K_{a} = \frac{[H_{3}O^{+}][IN^{-}]}{[HIN]} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}[/itex]


    3. The attempt at a solution

    Midway between the pH values for indicator HIN, or at pH 2.0, the hydronium ion concentration is equal to the Ka value. The hydronium ion concentration is [itex]10^{-2}[/itex]. The [itex]K_{a}[/itex] value is therefore [itex]10^{-2}[/itex].

    The same goes for the acid HA. Since at pH of 2.5, half of the acid has been ionized to hydronium ions, the strength of the acid, or Ka, is equal to [itex]10^{-2.5}[/itex].

    Therefore, through comparison of the Ka values, HIN is a stronger acid, as it has the larger Ka value.

    1) Is this correct? Is the train of thought also correct? I think I'm missing something here.
     
    Last edited: Dec 14, 2013
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  3. Dec 14, 2013 #2

    Borek

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    Staff: Mentor

    I have no idea what it is intended to mean. Is this how the question is worded? If so, perhaps someone else will have a go at it, my English fails me.
     
  4. Dec 14, 2013 #3

    Qube

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    Yes. I should clarify. HIN is an imaginary indicator and HA is an imaginary acid. HIN turns yellow if the solution has a pH below 1.0; green if the pH is 2.0, and so on.
     
  5. Dec 14, 2013 #4

    Borek

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    Sorry, I misquoted. The part I quoted earlier is perfectly clear, it is this phrase:

    that makes no sense to me.
     
  6. Dec 14, 2013 #5

    Qube

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    That's implying the pH is 2.5. HA is treated with indicator HIN and the color of the HA solution changes from colorless to green-blue.
     
  7. Dec 14, 2013 #6

    Qube

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    Let me explain how I re-did this problem. What I posted originally was incorrect. Now, I'm not sure what I'm gonna 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.

    [tex]HX (white) + H_{2}O \leftrightharpoons X^{-}(black) + H_{3}O^{+}[/tex]

    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, [itex]K_{a}[/itex], equals [itex]\frac{[H_{3}O^{+}][X^{-}]}{[HX]}[/itex].

    We can easily calculate [itex]K_{a}[/itex] understanding that the at pH of 8.0:

    [tex][H_{3}O^{+}] = [X^{-}] = x = 1 × 10^{-8}[/tex]

    and

    [itex][HX] = M_{i} - x[/itex]

    so

    [itex]K_{a} = \frac{[H_{3}O^{+}][X^{-}]}{[HX]} = \frac{x^{2}}{M_{i} - x}[/itex]

    and because we're at the halfway point, [itex]M_{i} - x = x[/itex], so

    [itex]K_{a} = \frac{[H_{3}O^{+}][X^{-}]}{[HX]} = \frac{x^{2}}{x} = x[/itex]

    thus,

    [itex]K_{a} = 1 × 10^{-8}[/itex]

    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:

    [itex]K_{a} (HIN) = 1 × 10^{-2}[/itex]

    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 [itex]K_{a} (HA)[/itex] to be able to compare the strengths of the two acids. We can use the framework I outlined above:

    [itex]K_{a} (HA) = \frac{[H_{3}O^{+}][A^{-}]}{[HA]} = \frac{x^{2}}{M_{i} - x} = \frac{x^{2}}{0.004 - x}[/itex]

    Since

    [itex]pH=−log[H_{3}O^{+}][/itex]

    it follows that

    [itex]10^{-pH} = [H_{3}O^{+}] = 10^{-2.5} = x [/itex]

    and thus:

    [itex]K_{a} (HA) = \frac{(10^{-2.5})^{2}}{0.004 - (10^{-2.5})} = 1 × 10^{-2}[/itex]

    In conclusion the strengths of the two acids are the same because we arrive at identical [itex]K_{a}[/itex] 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 [itex]K_{a} (HA)[/itex] value to one significant figure, and both acid ionization constants are identical to one significant figure.

    ----

    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!
     
    Last edited: Dec 14, 2013
  8. Dec 15, 2013 #7

    Borek

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    Assuming "treated" means just that the indicator was added to the acid solution (I would never thought about it is a "treatment" - perhaps that's my English failing me), your approach is correct.

    That is, you can easily delete 80% of the discussion, as it is not related to the problem.
     
  9. Dec 15, 2013 #8

    Qube

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    Yes, treated means the indicator was added to the acid solution in a very small quantity. And yes, the first part of my discussion is irrelevant; I was just trying to get things sorted out in my head. It is deeply satisfying to see that there are no misconceptions in what I posted!
     
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