pH electrodes

[vpena_1I]

For homework 6N.5(a), I'm good up until having
0.06V= -(0.0129 V) ln[H+]^2

How do I use this to find the pH? Im not sure how the solutions manual went from this step to
0.06V= -(0.0257 V) ln[H+]
= -0.0257V x (2.303 log[H+]
pH=1.0

[Connor Ho 1B]

The way the solutions manual does it is so that you are solving for -log[H+], which is equal to pH.

I think a simpler (or more understandable) way of getting to pH would be this:

Given: 0.06V= -(0.0129 V) ln[H+]²

Using log rules, you can bring the exponent down into the coefficient: 0.06V= -(0.0129 V)*2*ln[H+] = -0.0257V*ln[H+]

You can then divide so that you only have ln[H+] on the right side: -2.334 = ln[H+]

Cancel out the natural log by making each side of the equation an exponent of e: e^-2.334 = e^ln[H+]

This gives you: 0.09684 = [H+]

You then use the equation pH = -log[H+] to get: -log(0.096840) ≈ 1.0