Stoichiometry

[Stevensmay]

So even though the volume of water in the flask changed, the number of mole of vinegar itself hasn't. This means that it will still take the same amount of NaOH to neutralize the vinegar.
We have only diluted the vinegar, not changed how much of it there is.

[jack_chay]

you said the calculations will no longer be correct, but they said it would not affect the results?

[Stevensmay]

I wasn't sure what the question wanted. If we were to calculate concentration from our data, it would be wrong as we are using an incorrect volume.

The question does not seem to go that far, only to the mole calculation stage. This part will not change.

[jack_chay]

its ok thank you

[jack_chay]

how would I determine the pH of:
5 L of an aqueous solution that contains 1.0 grams of HBr and 1.0 grams of nitric acid.

Thanks in advance

[Stevensmay]

n(HBr) = 1/80.9
n(HNO3) = 1/63

1/80.9 + 1/63 = 1439/50967 mole of H+
Convert to a concentration, so 1439/254835 M H+
Put this into -log([H+]) and we get our answer, 2.2

I accidentally used the natural logarithm the first time Jack, sorry about that.

[jack_chay]

but the answer says 2.2?
what should I do?

[Stevensmay]

Above is now correct.
I was using a calculator I don't normally use for chem and forgot that it defaults to base e not 10.