pH = -log(H^+)
-3.25 = log(H^+)
Solve for (H^+). I estimated 5E-4 but you need to be more accurate than that.
..........HA ==> H^+ + A^-
I.........0.1.....0.....0
C.........-x.....x.....x
E......-0.1-x....x......x
and you know x = 5E-4 from above (again that is an estimate)
Ka = (H^+)(A^-)/(HA)
Solve for Ka.
Then .....HA ==> H^++ A^-
I........0.01.....0....0
C.........-x.....x......x
E......0.01-x.....x.....x
Substitute the E line into Ka expression along with Ka determined from the first part and solve for x = (H^+), then convert to pH.
The pH of a 0.10 M solution of a certain acid is 3.25 at 25°C. What is the pH of a 0.010 M solution of the same acid at the same temperature?
2 answers
Oh I see what I did wrong! I got the answer 3.75. Thanks for the help!
2 answers