Question

enough of a monoprotic acid is dissolved in water to produce a 1.51 m solution. the ph of the resulting solution is 2.85 . calculate the ka for the acid.

Answer 1

Ka = 1.32 x 10^-6

Step-by-step explanation:

First we should find the [H+].

pH = -log[H+], so [H+] = 10^-pH = 10^-2.85 = 0.00141 M

Then we can set up the equilibrium value

Which will be Ka = [A-][H+]/[HA], we can assume A- = H+

The final concentration of Acid will be initial - H+ as all H+ is formed from this acid.

Ka = [0.00141][0.00141]/[1.51-0.00141] = 1.32 x 10^-6