Respuesta :

Edufirst

Answer:

  • 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%

The percentage ionization is equal to 1.74%

Data;

  • concentration of benzoic acid = 0.215M

The Equation of Reaction

                [tex]HC_7H_5O_2 + H_2O \to C_7H_5O_2^- + H_3O^+\\[/tex]

initial           0.215           -             -                -

change        -x                               +x             +x

equilibrium   0.215 - x                    x                x

The equilibrium concentration of the acid

[tex]K_a = \frac{[C_7H_5O_2^-][H_3O^+}{[HC_7H_5O_2]}[/tex]

Let's substitute the value in this

[tex]K_a = \frac{x.x}{0.215-x}\\ K_a = \frac{x^2}{0.215-x} \\[/tex]

The Ka for benzoic acid is 6.5*10^-5

[tex]6.5*10^-^5 = \frac{x^2}{0.215 - x} \\[/tex]

since the value of Ka is very small.

0.215 - x = 0.215

[tex]x^2 = 6.5*10^-^5 * 0.215\\x = \sqrt{1.3975*10^-^5} \\x = 3.738*10^-^3[/tex]

[tex][H_3O^+] = x = 3.738*10^-^3[/tex]

The percentage ionization would be

[tex]\frac{[H_3O^+}{[HC_7H_5O_7]} * 100= \frac{3.738*10^10^-^3}{0.215} * 100 = 1.74 \%[/tex]

The percentage ionization is equal to 1.74%

Learn more on percentage ionization here;

https://brainly.com/question/15073874

https://brainly.com/question/9173942

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