Problem 14-79

To find the K_a of trichloroacetic acid:

	HCCl3CO2	⇌	H+	+	CCl3CO2−
I	0.050 M		0		0
C	0.050 M - x		+x		+x
E	0.050 M - x		x		x

Ka=	x2
	0.050 - x

HClO₄ is a strong acid, therefore it is completely dissociated in water
  0.040 M HClO₄ = 0.040 M [H⁺] in solution

Since both HCCl₃CO₂ and HClO₄ acid have the same pH, then [H⁺] are the same

Knowing that at equilibrium x = [H⁺] which is 0.040 M, then:

Ka=	(0.040)2	= 0.16
	0.050 - 0.040