Ka for the acid

Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.01850.0185 M solution. The pH of the resulting solution is 2.342.34. Calculate the K_a for the acid.

I keep getting 0.00113

Answer 1

This is pretty much the same question you posed earlier, and I answered earlier. Do it the same way.

Let the weak, monoprotic acid be represented by HA.

HA <==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

From the pH of 2.34, we can calculate the [H⁺], which is also the [A^-]

[H⁺] = 1x10^-2.34 = 4.57x10^-3 M

Ka = (4.57x10^-3)(4.57x10^-3) / 0.0185

Ka = 1.13x10^-3 (note: this Ka is not very small, so we should go back and adjust the [HA] as follows:

Ka = (4.57x10^-3)² / 0.0185 - 4.57x10^-3

Ka = 2.09x10^-5 / 0.01393

Ka = 1.50x10^-3