# Question afc26

Apr 22, 2015

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 $p {K}_{a}$ by using the acid dissociation constant

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(1.77 \cdot {10}^{- 4}\right) = 3.75$

The above equation becomes

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

$p {H}_{\text{sol}} = 3.85$

By definition, a solution's pH is equal to

$p {H}_{\text{sol}} = - \log \left(\left[{H}_{3} {O}^{+}\right]\right)$

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

$H C O O {H}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H C O {O}_{\left(a q\right)}^{-}$

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%")# 