It was determined that a 0.10 M solution of an acid was only 2.5% ionkzed. Find the Ka and pKa for the acid.

Answer 1

HA + H2O---> H3O+ + A-

Ka=[H3O+][A-]/[HA]

..........HA I H3O+ I A-

I.........I 0.1 M I 0 I 0 I
C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
E.........I 0.1-2(0.1*0.025) I (0.1*0.025) I (0.1*0.025)

Ka=[(0.1*0.025][(0.1*0.025]/[0.1-2(0.1*0.025]

pka=-log(ka)

Answer 2

Opps,

HA + H2O---> H3O+ + A-

Ka=[H3O+][A-]/[HA]

..........HA I H3O+ I A-

I.........I 0.1 M I 0 I 0 I
C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
E.........I 0.1-(0.1*0.025) I (0.1*0.025) I (0.1*0.025)

Ka=[(0.1*0.025][(0.1*0.025]/[0.1-(0.1*0.025]

pka=-log(ka)

I apologize about that one.