Question

Question #afc26

Answer 1

0.06%

Explanation:

The percent ionization of formic acid in your solution will be 0.0635%.

The first thing to notice is that you're dealing with a buffer, which is a solution that consists, in your case, of a weak acid, formic acid, and a salt of its conjugate base, sodium formate.

The great thing about a buffer is that you can use the Henderson-Hasselbalch equation to determine its pH.

`pH_"sol" = pK_a + log("[conjugate base]"/"[weak acid"])`

Calculate `pK_a` by using the acid dissociation constant

`pK_a = - log(K_a) = -log(1.77 * 10^(-4)) = 3.75`

The above equation becomes

`pH_"sol" = 3.75 + log(([HCOO^(-)])/([HCOOH])) = 3.75 + log((0.278cancel("M"))/(0.222cancel("M")))`

`pH_"sol" = 3.85`

By definition, a solution's pH is equal to

`pH_"sol" = -log([H_3O^(+)])`

This means that you can determine the concentration of hydronium ions by

`[H_3O^(+)] = 10^(-pH_"sol") = 10^(-3.85) = 1.41 * 10^(-4)"M"`

Formic acid dissociates according to the following equilibrium

`HCOOH_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + HCOO_((aq))^(-)`

The percent ionization of formic acid will be

`"% ionization" = ([H_3O^(+)])/([HCOOH]) * 100 = (1.41 * 10^(-4)cancel("M"))/(0.222cancel("M")) * 100`

`"% ionization" = color(green)("0.0635%")`