Question

Chloroacetic acid, HC2H2O2Cl, is a stronger monoprotic acid than acetic acid. In a 0.10 M solution, the pH is 1.96.

Calculate the pK_a for chloroacetic acid.

pK_a =

Answer 1

pH = -log [ H+ ]

-log[ H+ ] = 1.96

[ H+ ] = 1× 10^(-1.96 )

= 0.0109

HC2H2O2Cl <------> H+ + C2H2O2Cl

Ka = [ H+ ] [ C2H2O2Cl- ] / [ HC2H2OCl ]

[ H+ ] = [ C2H2O2Cl- ] = 0.0109M

[ HC2H2O2Cl ] = 0.1 - 0.0109M = 0.0891M

Therefore,

Ka = (0.0109)^2/0.0891= 1.33× 10^-3

pKa = -logKa

= - log(1.33×10^-3)

= 2.87