Question

Calculating concentration with pH...?

Answer 1

So If I have a pH pf 5, would the resulting concentration, with the formula pH=-log[H^+], become 100000? Because would would make my kA value, using kA=[H30+][A-]/[HA] really strange.

Answer 2

is this for real chemistry class or biology?

Answer 3

Well, its for a chemistry lab.

Answer 4

I was just doing some calculations, and then I realized I made a big mistake early on, this being the problem.

Answer 5

okay your chem lab would have a little bit of lesson
read and re-read it

Answer 6

Well, I did, but I think this is really an algebraic problem, much less a chemistry issue.

Answer 7

how did you determine the values or data?

Answer 8

For the pH, we were suppose to estimate it, using qualitative data.

Answer 9

from an actual experiment?

Answer 10

yea

Answer 11

hmmm pig HYPE

Answer 12

your formula for calculating pH is correct, but you plugged in your values incorrectly.\[pH = -\log[H^+] = -\log[1.0*10^{-5}]\]to reverse the function, we use the \(10^x\) function:
\[[H^+] = 10^{-pH}\]you dropped a negative sign somewhere, so the \(H^+\) concentration is 1*10^-5, not 1*10^5

Answer 13

Such a small would still work in the Ka and Kb calculations thought right.

Answer 14

um I think both ka and kb has to add up equaling 1*10^-14?

Answer 15

These are for different acids and bases, but you are right.

Answer 16

The Ka of an acid and the Kb of ITS CONJUGATE always multiply to 1*10^-14, which is the K of water. If you write the 2 dissociation reactions and add them together, you get the autoionization of water, which is where it comes from.

Answer 17

the small value of H+ still works in the Ka calculations, in fact it's quite large, when compared to other acids

Answer 18

No, no, these are 2 different acids and bases.

Answer 19

then their Ka values have nothing to do with each other