Question

Formic acid (HCOOH) is secreted by ants. Calculate [H₃O⁺] for a 4.98E-2 M aqueous solution of formic acid (K_a = 1.80E-4).

Answer 1

The dissociation constant for an acid shows the extent of its dissociation into protons and the respective conjugate base in a given medium. This parameter is essentially a equilibrium constant, which for formic acid is expressed as Ka = [H⁺][HCOO^-]/[HCOOH] in which the concentration of water is eliminated due to its abundance giving an equal presence in numerator and denominator. Now, from the Ka and concentration of the acid, using the expression for Ka, the concentration of protons can be found by rearranging the equation as [H⁺][HCOO^-] = [HCOOH] x Ka.

Since formic acid is monoprotic i.e. each molecule dissociates to give one to one ratio of protons and formate ion, the concentration of [H⁺] and [HCOO^-] will be equal. Thus, the expression boils down to x² = [HCOOH] x Ka where x is the concentration of protons and formate ions. So x is found as the square root of the product of Ka of formic acid and its concentration which is given by sq.rt of (4.98x10^-2)x(1.8x10^-4) = 2.994x10^-3M.

Thus the concentration of protons in the medium will be 2.994x10^-3M.