Question

A 0.2 m solution of a weak acid ha dissociates such that 99.4% of the weak acid remains intact (i.e., remains as ha). what is the pka of the weak acid?

Answer 1

HA -----> H(+) + A(-)
0,2-x.........x..........x

100% - 99,4% = 0,6% = 0,006

x = 0,6%×0,2 = 0,006 × 0,2 = 0,0012

HA --------------> H(+) + A(-)
0,2-x.....................x..........x
0,2-0,0012.....0,0012..0,0012
0,1988.............0,0012..0,0012

K = [H+][A-]/[HA]
K = (0,0012)^2/0,1988
K = 0,0000072434

pKa = -logK= -log0,0000072434 = 5,14

:)