The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

Answer 1

pH = 2.03 = -log(H^+). That will be about 0.01 but you need a more exact answer.
.................HF --> H^+ + F^-
I................0.01.....0.........0
C................-x........x..........x
E..............0.01-x....x..........x

Ka = (H^+)(F^-)/(HF). You know x, evaluate 0.01-x, plug into Ka expression and solve for Ka.
Post your work if you get stuck.

Answer 2

Can you solve it fully and explain it