Stuck on chemistry problem due tonight

Question

What is the pH and pOH of a solution that was made by adding 245 mL of water to 500 mL of 4 * 10 ^ - 4 M HF solution?

PH?round to 1 decimal place

pOH 1 decimal place

Answer 1

moles of HF = (4x10^-4) mole/L x 0.5 L = 2x10^-4 moles

molarity of final HF solution = 2x10^-4 moles/0.745L = 2.68 x10^-4 M

K_a of HF = 6.6 x10^-4

HF -----------> H⁺ + F^-

2.68 x10^-4 x x

Ka = [X] .[X] / [HF] ,,,,,,, 6.6 x10^-4 = X²/ 2.68 x10^{-4 ,} therefore X = [H⁺] = 4.20x10^-4

pH = -log( 4.20x10^-4) = 3.4; pOH = 14 - 3.4 = 10.6

Answer 2