Chem 102 pH Question

[OnMyWayThere]

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Hello... I just took my final in summer school for Chem 102 and this one problem is driving me crazy. It is worth 10% of the final and I'm not sure if I did it correctly. Can you please take a few minutes to see if I did it correctly?

We know E(cell) = E - .0592/n * logQ

.54 = .78 = .0592/2 * log 2/[H] ^2

(That's log 2 over [H] with just [H] squared)

And when we figure out [H], just do pH = -log [H] to get the answer.

I got 3.90 but when I asked some classmates, they got a very different result.

Thanks in advance...

J

 

[emomd]

check your answer, because i got something different.

 

[fun8stuff]

i think you would need to be a little more specific about what each number is where it came from and what the question was asking or me to fully answer your question.

one thing that you need to keep in mind is that perhaps the professor had more than one version of the test and that is why your classmates have different answers.

the question was asking for a pH, and you were given the E potentials? where did you get these numbers: .54 = .78? [H] is hydrogen concentration? how did you figure n was 2? perhaps n = something else? as for your math: i do not have a calculator on me and i'm just too lazy to do that log in my head...

 

[OnMyWayThere]

[H] is the Hydrogen concentration... The actual form is:

.54 = .78 - .0592/2 * log 2/ H^2


and then pH = - log [H]

Thanks...

 

[fun8stuff]

assuming your numbers and math are right, everything looks OK, but it is hard to say anything without A.) calculator, and B.) the actualy problem and where you got all those numbers from./

 

[mlw03]

i haven't done much math this whole summer, but i got [H]=0.193 and pH=.72