Approximately how much water should be added to 10.0 mL of 12.4 M HCl so that it has the same pH as 0.90 M acetic acid (Ka = 1.8 10-5)?

Answer 1

First you must calculate (H^+) for 0.90M acetic acid (HAc).
.........HAc ==> H^+ + Ac^-
I.......0.90.....0......0
C........-x......x......x
E.....0.90-x.....x......x

K = 1.8E-5 = (H^+)(Ac^-)/(HAc)
Plug in the E line and solve for x = (H^+).

So you want (H^+) in HCl to be the same.
10 mL x 12.4M = mL x MHAc from above.
mL then will be the total; subtract 10 from that to find how much must be added to 10. The reason the problem says "approximately" is because the total is not the sum of the two. Volumes are not additive although they would be so close in this instance that I expect we couldn't measure the difference.

Answer 2

You're dumb, you don't know how to explain the answer.

Answer 3

YOU DONOT KNOW ANY THING BROO