Question

A solution initially made up to be 0.260 M acid HX(aq) was prepared and the pH measured as 2.32. Calculate the pKa for the acid. Enter the pKa with 2 decimal places.

Answer 1

use:

pH = -log [H+]

2.32 = -log [H+]

[H+] = 4.786*10^-3 M

HX dissociates as:

HX -----> H+ + X-

0.26 0 0

0.26-x x x

Ka = [H+][X-]/[HX]

Ka = x*x/(c-x)

Ka = 4.786*10^-3*4.786*10^-3/(0.26-4.786*10^-3)

Ka = 8.976*10^-5

use:

pKa = -log Ka

= -log (8.976*10^-5)

= 4.0469

Answer: 4.05