What is the pH of a 1.5 M solution of trifluoroacetic acid (CF3CO2H) with pKa = 0.2?

Trifluoroacetic acid is a weak acid, which means it partially dissociates in water to produce hydrogen ions (H+) and trifluoroacetate ions (CF3CO2-):

{eq}CF_3CO_2H rightleftharpoons H^+ + CF_3CO_2^- {/eq}

The equilibrium constant expression for this reaction is:

{eq}K_a = dfrac{[H^+][CF_3CO_2^-]}{[CF_3CO_2H]} {/eq}

where [H+] represents the concentration of hydrogen ions, [CF3CO2-] represents the concentration of trifluoroacetate ions, and [CF3CO2H] represents the concentration of undissociated trifluoroacetic acid.

The pKa of trifluoroacetic acid is given as 0.2, which means:

pKa = -log Ka

0.2 = -log Ka

Ka = 10^-0.2

Ka = 0.01

Now, we can use the equilibrium constant expression and the known value of Ka to calculate the concentration of hydrogen ions in a 1.5 M solution of trifluoroacetic acid:

Ka = [H+][CF3CO2-]/[CF3CO2H]

0.01 = [H+]^2/[CF3CO2H]

[H+]^2 = 0.01 x 1.5

[H+]^2 = 0.015

[H+] = sqrt(0.015)

[H+] = 0.1227 M

Therefore, the pH of the solution can be calculated using the formula:

pH = -log[H+]

pH = -log(0.1227)

pH = 0.912

So, the pH of a 1.5 M solution of trifluoroacetic acid with a pKa of 0.2 is \approx imately 0.912.