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### Topic: Electrode Repsponse Question  (Read 3122 times)

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#### Halogen876

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« on: August 14, 2014, 02:29:39 PM »
Hello,

I've been attempting to work out the following question and have run into some problems. Here is the question:

You are unsure about the ideality of response of a liquid membrane electrode containing an ion carrier responsive to K+. You therefore make the following measurements: EISE=0.1131V in 0.00800M KCl, EISE=0.0824V in 0.00200M KCl. Just before you are to measure your unknown, a fellow student breaks the reference electrode and ruins the 0.00800M standard solution. Another reference electrode is found (but you are not sure whether it is the same kind as the one that was broken). The voltage of the ISE in your unknown is 0.0237V vs. the new reference and the voltage of a mixture of 10.00mL of your unknown and 5.00mL of the 0.00200M KCl solution standard is 0.0318V. What is the concentratin of potassium in your unknown?

I attempted to solve this problem by ignoring the measurements that were made on the original electrode and setting up 2 equations for the 2 measurements made on the new electrode and subtracting those equations to solve for [K+] I was able to get an answer but it wasn't the correct answer. (The correct answer is 8.60E-4M.) I must be overlooking something, but I can't think what it is, so if anyone has any ideas, I'd really appreciate any help #### Borek ##### Re: Electrode Repsponse Question
« Reply #1 on: August 14, 2014, 03:37:40 PM »
I don't think you should ignore the first two measurements. I would try using them to find the slope.
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#### Halogen876

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« Reply #2 on: August 14, 2014, 05:29:27 PM »
Ok, so using the first 2 measurements, the slope would be (0.1131V-0.0824V)/(0.00800M-0.00200M)=5.12V/M.

I'm not quite sure where to go with that now though, since the second electrode might be different so I don't know that the same slope would be applicable - so any more insight would be greatly appreciated! Thanks!

#### mjc123

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« Reply #3 on: August 15, 2014, 04:43:50 AM »
That is the wrong calculation - voltage does not vary linearly with concentration. Do you know the equation linking voltage to concentration?

The same slope should be applicable, as that is a property of the ISE, not the reference electrode. A different reference might have a different voltage, meaning the calibration curve would be shifted, but still have the same slope. That's why a single measurement can't give you your answer.

#### Halogen876

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« Reply #4 on: August 15, 2014, 10:26:52 AM »
I think the equation I should be using is the Nernst equation, so E=E0-0.0592/nxlog[K+].

E0 could be found in a table and n is 1. I'm not quite sure how to get a slope out of that though...

I think I understand what you mean about how the slope would be the same using different refernce electrodes, so can I disregard the fact that the measurements are being taken on 2 different electrodes?

I think I'm starting to make a little sense of this, but it's all still pretty cloudy to me so any more suggestions would be appreciated! Thanks!

#### mjc123

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« Reply #5 on: August 15, 2014, 12:27:57 PM »
The Nernst equation gives the ideal situation; in reality it doesn't always follow Nernst perfectly, which is why the question began "You are unsure about the ideality of the response..." In practice you measure the slope of the voltage curve: E1 - E2 = S*log(C1/C2) where the slope S is ideally 0.059/n, but in fact may be a bit different.
To answer the question you need to assume that the slope is the same with the second reference electrode as with the first. Work out expressions for the concentrations in the two solutions and plug them into the above equation.

#### Halogen876

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« Reply #6 on: August 15, 2014, 03:50:44 PM »
That makes sense now - I got it figured out! Thank you both so much for all your help 