Enough of a monoprotic weak acid is dissolved in water to produce a 0.0102 M solution. The pH of the resulting solution is 2.68 . Calculate the for the acid. K_{a}

Answer 1

Let HA be the monoprotic acid.

HA ==> H⁺ + A^-

pH = -log [H⁺]

2.68 = -log [H⁺]

[H⁺] = [A^-] = 2.09x10^-3 M

Ka = [H⁺][A^-] /[HA]

Ka = (2.09x10^-3)(2.09x10^-3) / 0.012

Ka = 3.64x10^-4

For more advanced chemistry course:

Ka = [H⁺][A^-] /[HA] - [H⁺]

Ka = (2.09x10^-3)(2.09x10^-3) / 0.012 - 2.09x10^-3

Ka = 4.41x10^-4