What is the pH of 0.180 M solution of hydrofluoric acid? (Ka = 6.6 x 10-4)

Question

What is the pH of 0.180 M solution of hydrofluoric acid? (Ka = 6.6 x 10-4)

	A. 0.745
	B. 0.180
	C. 0.090
	D. 1.96

Answer 1

HF ==> H+ +F-

Ka = 6.6x10^-4 = [H+][F-] / [HF] = (x)(x) / 0.180

x² = 1.12x10^-4

x = 1.09x10^-2 M = [H⁺]

pH = -log 1.09x10^-2

pH = 1.96 (Answer D)