Question #afc26

1 Answer

Answer:

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%")#

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