Enough of a monoprotic weak acid is dissolved in water to produce a 0.0110 M solution. The pH of the resulting solution is 2.34 . Calculate the Ka for the acid.

Answer 1

Let the monoprotic acid be HA.

HA <==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

From the pH, we can find the [H⁺] and [A^-] and we are given the [HA] as 0.0110 M. We can now find Ka.

pH = 2.34, therefore [H⁺] = [A^-] = 1x10^-2.34 = 4.57x10^-3 M

Ka = (4.57x10^-3)(4.57x10^-3)/0.0110

Ka = 1.9x10^-3