In a lab the task was to dilute a 1.0M sol'n of Acetic Acid to a sol'n with a final pH of 4.15.

Question

The method I used to solve this was apparently wrong, but I don't understand how or why. Please help!

Here is what I did:
AcOH + H2O <-> H3O+ + AcO-

First I found the final desired [H+]
[H+]= 10^-pH= 10^-4.15 = 7.1x10^-5

Then I diluted my initial AcOH sol'n to 0.01M -> add 90mL H2O to 10mL AcOH. Do this step 2x.

Next I used Ka to calculate the [H+] at Equilibrium in the initial 0.01M solution.
Ka= [H+][AcO-] Ka= [x][x]
---------- -------
[AcOH] 0.01-x

[H+]eq = 4.2x10^-4(assume 0.01-x = 0.01)

Then, to achieve pH 4.15, you need to dilute using the following formula:
[H+]f = [H+]i x (Vi/Vf)
7.1x10^-5 = 4.2x10^-4 x (10mL/ ?mL)
Vf = 59.15mL 59.15mL-10mL = 49.15mL water needed to dilute to pH 4.15.

I'm not sure where I went wrong here. Please advise! Thank you!! :)

Answer 1

You must realize that you can't calculate the (H^+) and dilute to that mark BECAUSE dilution of the HOAc will ionize the acetic acid more (remember Le Chatelier's Principle)
AcOH + H2O ==> H3O^+ + AcO^-
and adding water shifts the equilibrium to the right which changes the (H3O). In addition, when working with such small concns I like to carry the calculation to more places (even more than I know I'm allowed) so as to prevent rounding errors, then round at the end. I would do this.
pH = -log(H^+) and (H^+) = 7.079E-5 and plug this into the Ka expression and solve for the (AcOH). It's the AcOH you want to have as some final concn. That will not change much with the addition of water.
Ka = 1.75E-5 =(7.079E-5)^2/[(AcOH)-7.079E-5] and solve for (AcOH).
Then I would take 1 mL of the 1.0 M stuff and dilute to the (AcOH). I did a quick calcn and if I didn't slip up the concn AcOH you want is 3.57E-4M. I solved the quadratic both times, once when finding (AcOH) and once when I take that value and work backwards to see if I come out with pH 4.15. I did.

Answer 2

Thank you!