About Using ln or log


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Yinhan_Liu_1D
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About Using ln or log

Postby Yinhan_Liu_1D » Thu Feb 09, 2017 3:25 pm

Hi, dear all,

I find it confusing using the equation given in ln and the equation given in log.

For example in HW 14:97:

If we use ln: lnK=nE/0.02569v = -12 -->K=6.14*10^-6
But if we use log as in the course reader: logK= nE/0.0592=-5.4 --> K= 3.93*10^-6

And by taking the square root of them, I get different answers of Ka. According to the solution manual, the one using ln is right.

Did I miss something in the Calculation? Anything wrong? I would really appreciate your help!

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Re: About Using ln or log

Postby Chem_Mod » Thu Feb 09, 2017 4:41 pm

and are both extremely small numbers and are actually very close in value. You can use either the log or ln of the nernst equation, but the ln version is more accurate, because the log equation presents (RT/F) as a constant, and so it is rounded and not as accurate. Also, the constant in the log version is calculated using a temperature of 298 K (25 degrees Celsius), so you cannot use that constant when temperature is not 298 K.


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