Consider 51.5 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 4.210. What volume of water must be added to make the pH = 5.000

Question

I've tried everything but I am not getting the right answer!

Answer 1

I would do this.
........HA --> H^+ + A^-
I.......x......0.....0
C......-y......y.....y
E .....x-y.....y.....y

You know y = 6.17E-5
Substitute and solve for x =initial (HA) and let's call that M1. (This will be quadratic you must solve.)

Do the same for HA in which y = 1E-5 and solve for HA and let's call that M2.

Then 51.5 x M1 = mL x M2. (another quadratic)
Solve for mL which will be the final volume, then mL-51.5 will be how much H2O must be added.