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Topic: Problem of the week - 14/05/2012  (Read 6188 times)

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Online Borek

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Problem of the week - 14/05/2012
« on: May 14, 2012, 09:19:38 AM »
18 test tubes were put in two rows in a test tube rack. 10 mL of 50 mM HCl were put into each test tube in the front row, 10 mL of 0.1M NaOH were put into each test in the back row. Tubes in the front row were marked 1, 2, 3... up to 9. To the tube marked 1 one drop of the methyl red solution was added, two drops were added to the test tube marked 2, three drops to the test tube marked three and so on, up to nine drops in the front row test tube marked nine. Then, the same solution of the methyl red was added to NaOH filled test tubes from the back row, but in reversed order - one drop to the test tube behind the one marked 9, two drops to the test tube behind the one marked 8 and so on, up to nine drops put in the test tube behind the one marked 1. Source of a white light was put behind the rack so that it was possible to observe hue of the light after passing through both solutions.

Two additional test tubes were filled one with a pure water and the other one with unknown solution. To the one with unknown solution 10 drops of the same methyl red solution were added. When these two tubes were observed in identical conditions, observed hue was almost identical to the one observed when the light was passing through the test tubes marked with digit 3.

What is pH of the unknown solution, if pKa of the methyl red is 5.0?
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Offline cth

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Re: Problem of the week - 14/05/2012
« Reply #1 on: May 18, 2012, 09:16:17 AM »
The acid solution has a pH of:  - log(0.05) = 1.3
The basic solution has a pH of: 14 + log(0.1) = 13


We have nine series of test tubes containing known solutions, but we can imagine two more:
- test tube series marked 0 which consist in 10 drops of methyl red in the basic solution at pH 13;
- test tube series marked 10 which consist in 10 drops of methyl red in the acidic solution at pH 1.3;


The eleven series of test tubes cover a range of pH from 1.3 to 13.
So between two adjacent series, the pH variation is: (13 - 1.3) / 10 = 1.17

So, test tubes marked 3 have a pH of: 13 - 3*1.17 = 9.49

The unknown solution has a pH around 9.5

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Re: Problem of the week - 14/05/2012
« Reply #2 on: May 18, 2012, 09:27:33 AM »
No.

Think about it - can you use a single indicator to measure pH that is 4.5 pH unit from the indicator pKa?
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Offline Yggdrasil

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Re: Problem of the week - 14/05/2012
« Reply #3 on: May 19, 2012, 10:52:50 PM »
In the acidic tube, we have the protonated form of the indicator (HA) while in the basic tube we have the deprotonated form of the indicator (A-).  Because the unknown solution has the same color as tube pair #3, this means that the ratio of protonated indicator to deprotonated indicator ([HA]/[A-]) is 3:7.

Using the Henderson-Hasselbalch equation, we can calculate the pH of the solution:
pH = pKa - log([HA]/[A-] = 5.0 - log(3/7) = 5.4

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Re: Problem of the week - 14/05/2012
« Reply #4 on: May 20, 2012, 04:23:36 AM »
Yes, that's the correct approach and the correct answer.

I told you it wasn't difficult :)

Based on COLORIMETRIC DETERMINATION OF TITRATION CURVES WITHOUT BUFFER MIXTURES, L.J. Gillespie, J.Am.Chem.Soc. 42, 744 (1920).
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