Question

The Ka of formic acid is 1.77 ⋅ 10-4. Calculate the percent ionization of formic acid (HCO2H) in a solution that is 0.125 M in formic acid. The Ka of formic acid is 1.77 10-4. 0.859 0.0180 3.79 2.25 ⋅ 10-5 6.94

Answer 1

Answer

The percent ionization of formic acid is 3.69%.

Explanation

Given:

Ka of formic acid = 1.77 x 10⁻⁴

Molarity of formic acid = 0.125 M

What to find:

The percent ionization of formic acid.

Step-by-step solution:

The dissociation of formic acid is given as:

$HCO_2H\rightleftarrows H^++HCOO^-$

The acid dissociation constant (Ka) for formic acid is given as:

$K_a=\frac{[H^+][HCOO^-]}{[HCO_2H]}$

Substituting the concentration of the ions and the acid into the acid dissociation constant above:

$\begin{gathered} 1.77\times10^{-4}=\frac{[x][x]}{[0.125-x]} \\ \\ x=4.61\times10^{-3} \end{gathered}$

The hydrogen ion concentration in the solution = 0.00461 M.

The percent ionization of the formic acid can be calculated using the formula below:

$Percent\text{ }ionization=\frac{[H^+]}{[HCO_2H]}\times100\%$

Putting [H⁺] = 0.00461 M and [HCO₂H] = 0.125 M into the formula

$Percent\text{ }ionization=\frac{0.00461}{0.125}\times100\%=3.69\%$

The percent ionization of formic acid is 3.69%.