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Topic: ISE problem- need a little guidance!  (Read 2014 times)

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Offline gracelee

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ISE problem- need a little guidance!
« on: December 08, 2014, 11:52:27 AM »
This is problem that I have been having a really hard time with.  And I need to figure it out before upcoming exam, finals, etc. in case it shows up! Basically- you have a solution containing both Ca2+ and Mg2+ which you titrate with EDTA at pH 10, from this you find that the combined concentration of [Ca2+ + Mg2+] = 0.01523M (that part was easy.) Next, you take the same solution, dilute by factor of 20 with KN03, and measure the resulting solution with a calcium-ion selective electrode, getting a voltage of -41.4mV. Meanwhile...you have already created a standard curve for the calcium ion selective electrode using standard solutions of CaNO3. The equation of the best-fit line is (y = 29.6x + 80.4). (fyi- the curve is mV vs. Log[Ca2+]). The electrode's selectivity coefficient is KCa2+, Mg2+ = 0.00060. So..we want to find the individual concentrations of [Ca2+] and [Mg2+] in the original solution.

My thinking-- we have the concentration [Mg2+ + Ca2+]. My immediate thought is plug the voltage, -41.4mV, into the standard curve equation and determine [Ca2+]. Then subtract it from the concentration of [Ca2+ + Mg2+]. But this obviously doesn't take into account the fact that the measured voltage, -41.4mV is not just a response to Ca2+, it's the electrode's response to both Mg2+ and Ca2+!

Possible methods of solution --> I'm thinking if I could use the selectivity coefficient to determine the error in the voltage due to Magnesium (but I'm not sure how I would do that)- then I could use the error-adjusted voltage to find [Ca2+], using the standard curve, and then subtract that value from the combined concentration of Mg2+ and Ca2+.
OR, I'm wondering if maybe I need to use some kind of system of simultaneous equations?  I could do this but I'm still hung up on the fact that I think the measured voltage, -41.4mV, as it is, shouldn't be used to determine Ca2+ from the standard curve because it's not all due to Ca2+!
Any help/guidance with this would be greatly appreciated! 

Offline Borek

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Re: ISE problem- need a little guidance!
« Reply #1 on: December 08, 2014, 12:09:47 PM »
Simultaneous equations is a way to go IMHO. Can you write an equation that shows how the ISE responds to both ions present? It will be a variant of the Nernst equation, just using both concentrations and the selectivity coefficient.
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Offline gracelee

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Re: ISE problem- need a little guidance!
« Reply #2 on: December 08, 2014, 12:56:24 PM »
Thank you for your reply!

Okay, so equations would be:

E = Constant + (0.05916/2)Log([Ca2+] + KCa2+, Mg2+[Mg2+])

and

[Mg2+ + Ca2+] = 0.01523

I kept getting hung up on the standard curve...but maybe the purpose of it is just so the constant can be calculated?

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