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Topic: Electrochemistry - fluoride electrode  (Read 1577 times)

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Offline well

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Electrochemistry - fluoride electrode
« on: December 17, 2013, 11:30:12 AM »
A fluoride electrode is used to determine fluoride in a 10-mL sample, The sample is diluted to 100mL with TISAB solution and measured with the electrode. The reading obtained is -226.5mV.
If a 2mL of 5.00 x 10^-2 M fluoride standard is added to the original sample and the solution is -238.6mV. If Nernstian response is assumed, what is the concentration of fluoride in the original sample?


I guess:
No. of mole of  fluoride standard = (5.00 x 10^-2) x (2 x 10^-3) = 1 x 10^-4 mol
Since E = Q - 0.0592 log [F-],

0.2265 = Q - 0.0592 log[ ((1 x 10^-4) + (No. of mole of F- in sample) / (100 x 10^-3)]
0.2386 = Q - 0.0592 log[ (No. of mole of F- in sample)/ (100 x 10^-3)]

Is this flow correct? However, I do not how to solve these equations.
« Last Edit: December 17, 2013, 01:01:05 PM by well »

Offline Borek

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Re: Electrochemistry - fluoride electrode
« Reply #1 on: December 17, 2013, 04:18:05 PM »
0.2265 = Q - 0.0592 log[ ((1 x 10^-4) + (No. of mole of F- in sample) / (100 x 10^-3)]
0.2386 = Q - 0.0592 log[ (No. of mole of F- in sample)/ (100 x 10^-3)]

Looks OK to me.

Quote
However, I do not how to solve these equations.

Simple math - two unknowns, two equations.

See if subtracting second from the first won't remove one of the unknowns.
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