Consider 25.0 mL of 0.100M HA (HA is a fictional acid), for which Ka=1.00x10^-5, titrated with 0.0500M NaOH solution. What is the initial pH before adding any titrate?

Answer 1

................HA ===> H^+ + A^-
I...............0.100........0........0
C................-x.............x........x
E...........0.100-x..........x........x

Ka = (H^+)(A^-)/(HA) = 1E-5
Substitute the E line into the Ka expression and solve for x = (H^+), then convert to pH. pH = -log (H^+)
Post your work if you get stuck.

Answer 2

I know how to do the pH afterwards but what's confusing me is that sub into the E line part? Like;

1.00x10^-5 = (H^+) (what goes here though) / (what value goes here?)

How can I find the other values for A- and HA to fill in so I can solve for H+?