What volume (mL) of .20 M HCL needs to be added to 1.23 L of .10 M acetic acid to reduce percent ionization to 1.00%?

Answer 1

I would do this.
........HAc ==> H^+ + Ac^-
I.......0.1......0.....0
C........-x......x.....x
E.....0.1-x......x......x

Plug the E line into the Ka expression and solve for x = (H^+). I obtained 1.34E-3M for (H^+)

.........HAc ==> H^+ + Ac^-
What you want to do now is to add H^+ from the HCl so that the equilibrium will shift to the left and that decreases the ionization. So how much HCl must we add.
1.8E-5 = (H^+)(Ac^-)/(HAc)
If you want (Ac^-)(HAc) to be 0.01, then (H^+) must be what? That's
(H^+)(0.01) = K = 1.8E-5
and H^+ = 0.0018M
It was already 0.00134 so we must add how much? That's 0.0018-0.00134 = 4.6E-4M. You want 1.23 L of that so 4.6E-4M x 1.23 L = 5.66E-4 mols and since M = mol/L then L = mols/M =- 5.66E-4/0.2MHCl = 2.83E-3 L or 2.83 mL of the 0.2M stuff. You can check that out to see if it gives a solution that is 1% ionized.
That 0.0018M H^+ will be slightly diluted by 0.00180 x (1230/1232.8) = ?
The 0.1M HAc will be diluted by the same proportion of 0.1 x (1230/1232.8) = ?
Plug those values into the Ka expression and solve for (Ac^-), then
% ion = 100*(Ac^-)/(HAc). I came up with 0.99% which is about as close as I can get.

Answer 2

I made a typo part way down the page. Instead of retyping the whole things, since it is so long, here is the line in bold. I show it as multiplied and it should be divided.
If you want (Ac^-)/(HAc) to be 0.01, then (H^+) must be what?

Answer 3

I re-thought that problem (all night long) and there is nothing wrong with the first part; i.e., (H^+) must be 0.0018 to make it work and you must add 2.83 mL of the 0.2 M acid to get that. When I tried to check it to see if all that worked the best I could do was 0.99% and that wasn't good enough. Here is how it is checked.

...........HAc ==> H^+ + Ac^-
I..........0.1...0.00134..0.00134
add..............4.6E-4MHCL.....
C...........+x..............-x
E.........0.1+x....0.0018..0.00134-x

Set up Ka = (H^+)(Ac^-)/(HAc) and solve for x = 3.37E-4M. That makes Ac^- = 0.00134-3.37E-4 = 0.001
Then % ion = (0.001/0.1)*100 = 1.00%.
So it checks out.