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

The percent ionization of formic acid is 3.69%.

Step-by-step 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=([H^+][HCOO^-])/([HCO_2H])`

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

`\begin{gathered} 1.77*10^(-4)=([x][x])/([0.125-x]) \\ \\ x=4.61*10^(-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=([H^+])/([HCO_2H])*100\%`

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

`Percent\text{ }ionization=(0.00461)/(0.125)*100\%=3.69\%`

The percent ionization of formic acid is 3.69%.