Ka=x^2/.02
x=antilog -3.31= you do it.
then square it, divide by .02
What is Ka for 4-aminobenzoic acid if a 0.020 M solution of the acid has a pH of 3.31?
a. 2.5 ´ 10-2
b. 2.0 ´ 10-2
c. 4.9 ´ 10-4
d. 1.2 ´ 10-5
e. 2.8 ´ 10-6
1 answer
1 answer