The degree to which a weak acid dissociates is given by the acid -ionization con

Question

The degree to which a weak acid dissociates is given by the acid-ionization constant, Ka . For the generic weak acid, HA

HA(aq)+H2O(l)A(aq)+H3O+(aq)

this constant is given by

Ka=[A][H3O+][HA]

Strong acids will have a higher value Ka. Similarly, strong acids will have a higher percent ionization value.

Percent ionization=[A] equilibrium[HA] initial×100%

Strong acids, for which is Ka very large, ionize completely (100%). For weak acids, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization.

Benzoic acid, C6H5COOH, is a weak acid with a Ka value of 6.5×105.

Part A

What is the pH of a 0.455 M  benzoic acid solution?

Express your answer numerically to two decimal places.

Part B

What is the percent ionization of benzoic acid at this concentration?

Express your answer with the appropriate units.

The degree to which a weak acid dissociates is given by the acid-ionization constant, Ka . For the generic weak acid, HA

HA(aq)+H2O(l)A(aq)+H3O+(aq)

this constant is given by

Ka=[A][H3O+][HA]

Strong acids will have a higher value Ka. Similarly, strong acids will have a higher percent ionization value.

Percent ionization=[A] equilibrium[HA] initial×100%

Strong acids, for which is Ka very large, ionize completely (100%). For weak acids, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization.

Benzoic acid, C6H5COOH, is a weak acid with a Ka value of 6.5×105.

Part A

What is the pH of a 0.455 M  benzoic acid solution?

Express your answer numerically to two decimal places.

Part B

What is the percent ionization of benzoic acid at this concentration?

Express your answer with the appropriate units.

Explanation / Answer

               C6H5COOH --------------------> C6H5COO- + H+

I              0.455                                        0                   0

C             -x                                             +x                   +x

E          0.455-x                                        +x                   +x

               Ka   = [C6H5COO-][H+]/[C6H5COO-]

                6.5*10-5   = x*x/0.455-x

                 6.5*10-5 *(0.455-x) = x2

                      x    = 0.0054

           [H+] = x = 0.0054M

          PH    = -log[H+]

                  = -log0.0054   = 2.2676

Percent ionization=[H+] equilibrium/[C6H5COOH] initial×100

                             = 0.0054*100/0.455   = 1.186%

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