When an ion-selective electrode for X was immersed in 0.0600 M XCl, the measured

Question

When an ion-selective electrode for X was immersed in 0.0600 M XCl, the measured potential was 0.0430 V. What is the concentration of X when the potential is 0.0580 V? Assume that the electrode follows the Nernst equation, the temperature is at 25°C, and that the activity coefficient of X is 1.

So basically, I know how to solve this up until the last step where I calculate the concentration of [X+] in the Nerst equation. I'm not sure how to solve it step by step or how to enter it into my calculator.

Here is what I have:

E = Ecell + (8.314 x 298.15)/(1 x 96485) x ln[X+]

E = Ecell + 0.02569 ln[X+]

when [X+] = 0.0600 and E = 0.0430

0.0430 = Ecell + (8.314 x 298.15)/(1 x 96485) x ln(0.0600)

Ecell = 0.1153 V

when E = 0.0580 V

0.0580 = 0.1153 + (8.314 x 298.15)/(1 x 96485) x ln[X+]

How do I solve for [X+] here, step by step? How do I enter it into a calculator?

Explanation / Answer

Every step whatever it has given as an answer it is correct. Except for simplification of the last step.

0.0580 = 0.1153 + (8.314 x 298.15)/(1 x 96485) x ln[X+]

ln [X+] = 2.303 log [X+ ] = -2.23 [From logarithm relations]

log [X+] = - 2.23 /2.303 = - 0.968

[X+] = 10-0.968 = 0.107 M

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