Determine the percent ionization of a 0.215 m solution of benzoic acid.

1.69%

Explanation:

The percent ionization of a weak acid is the percent of the original acid that ionizes.

% ionization = [ concentration of the ion at equilibrium / original concentration of acid ] × 100

You may calculate the percent ionization of an acid of known concentration, from the equilibrium constant.

So, to determine the percent ionization of a 0.215 M solution of benzoic acid, you must look for the equilibrium (dissociation or ionization) constant.

Equilibrium constants depend on temperature. So, you must know the temperature.

For this question, I will assume 25°C, for which you can find that the dissociation constant, Ka, for benzoic acid is 6.25×10⁻⁵.

With that, you follow these steps:

1. Write the ionization equation:

C₆H₅CO₂H ⇄ C₆H₅CO₂⁻ + H⁺  (simplified version)

2. Calculate the concentration of the ion C₆H₅CO₂⁻ at equilibrium

ICE (initial, change, equilibrium) table:

                    C₆H₅CO₂H ⇄ C₆H₅CO₂⁻ + H⁺

       I           0.215                  0              0

      C            - x                   + x           + x

      E          0.215 - x              x              x

     

 Equilibrium expression:

        Ka = x² / 0.215 - x =  6.25×10⁻⁵

 Solve for x (assume 0.215 >> x)

           x² = 0.215 × 6.25×10⁻⁵ = 0.0000078125

           x = 0.00367 M

           If you do not make the assumption but solve the quadratic equation you will get x = 0.00363 M

3. Calculate the percent ionization:

With x = 0.00363 M (exact calculation)

% = [0.00363 / 0.215] × 100 = 1.69%

With x = 0.00367 M

% = [0.00367 /0.215] × 100 = 1.71%

Hi; /c

The answer for this is a