Question

How do I solve this kind of problem?

What is the pH of 0.010 M aqueous hypochlorous acid? (Ka of HOCl=3.5*10^-8)

Answer 1

Here is a rough sketch of the steps:

1. Set up the ICE table for the equilib:

HOCl + H2O <=> OCl^- + H3O^+

the initial conc of HOCl is 0.010 M. let's say x M of HOCl dissociates, then the equilib conc becomes 0.01-x M. (need more help?)

2. At equilib:

[OCl-] [H3O+] / [HOCl] = Ka = 3.5*10^-8

(need more help here?)

3. the rest is solving for the unknown x. x is the equilib. conc of H3O+, and the pH is the negative log (base 10) of x.

(need more help here?)

Answer 2

How would I set up the ICE for this one? I can't figure out the equilibrium values?

Answer 3

well, the ICE table's sole purpose is to help us express write the equilib. values in terms of the unknown variables.

let's say that at equilibrium, x M of 0.010 M of HOCl has donated the proton to water. then we would have the following equil. values:

HOCl: 0.01-x M
H3O+: x M
OCl-: x M

agree?

Answer 4

Yes, and x ends up being 1.87*10^-5?

Answer 5

i have pretty much the same value by approximating 0.01-x with 0.01. sometimes we have to apply the quadratic formula, but in this case 10^-5 is less than 5 percent of 0.01 so it's okay.

do you know how to get the pH from x?

Answer 6

Well, I originally thought that I'd take the -log of OH- and subtract it from 14 to get the pH, but that doesn't seem to be working.

Answer 7

x is the equilib. conc of H3O or really just "H+" for our purpose, so pH is the -log of x. agree?

Answer 8

Agreed, so the final pH would be ~4.73?

Answer 9

that's what i have. and it makes sense because HOCl is an acid. so the pH of an HOCl solution will be lower than 7. it's a weak acid, so the pH is not very low like 1 or 2.

Answer 10

looks like i helped, awesome!

Answer 11

Great explanations, thanks for all the help! You rock :)