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E(observed) = E(cell) -(.06)*log[H+]
If Ecell were actually .3V rather than .285V what would be true of the pH measured, assuming no correction is made?
The answer: The pH determined by the pH meter would be too low, because the Ecell in calculations is too high.
How do they reason this?
Oops. Solved it, I think.
I re-wrote it as E(observed) = E(cell) +(.06)*pH
E(observed) = Constant (is this the right assumption?)
E(cell)= Big or E(cell)= Small
pH = -log[H+] (can ignore the .06)
So Constant = 3 + pH vs Constant = 2 + pH
thus, its the condition where E(cell) was large, pH would have to be smaller.
Is there a quicker, more intuitive way?
If Ecell were actually .3V rather than .285V what would be true of the pH measured, assuming no correction is made?
The answer: The pH determined by the pH meter would be too low, because the Ecell in calculations is too high.
How do they reason this?
Oops. Solved it, I think.
I re-wrote it as E(observed) = E(cell) +(.06)*pH
E(observed) = Constant (is this the right assumption?)
E(cell)= Big or E(cell)= Small
pH = -log[H+] (can ignore the .06)
So Constant = 3 + pH vs Constant = 2 + pH
thus, its the condition where E(cell) was large, pH would have to be smaller.
Is there a quicker, more intuitive way?
Last edited:
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