What is the acid dissociation constant for a 0.0250 M monoprotic acid when [H+] = 4.5x10^-5 ?

Answer 1

Let HA be the weak acid.

HA <==> H⁺ + A^-

Keq = [H⁺][A^-]/[HA] = (4.5x10^-5)(4.5x10^-5)/0.025 - 4.510^-5

Since 4.5x10^-5 is very small compared to 0.025, we can ignore it in the denominator of the above equation.

Keq = (4.5x10^-5)²/0.025

Keq = 8.1x10^-8