In addition to the acid-dissociation constant, K_a, another characteristic of so

Question

In addition to the acid-dissociation constant, K_a, another characteristic of solutions of acids is percent ionization, determined by the following formula: Percent ionization = [HA]ionized/[HA] initial times 100% Percent ionization increases with increasing K_a. Strong acids, for which K_a is very large, ionize completely (100%). For weak acids, the percent ionization changes with concentration. The more diluted the acid is, the greater percent ionization. A convenient way to keep track of changing concentrations is through what is often called an I.C.E table, where I stands for "Initial Concentration, " C stands for "Change, " and E stands for "Equilibrium Concentration." To create such a table, write the reaction across the top, creating the columns, and write the rows I.C.E on the left-hand side. A + B rightarrow AB Initial (M) Change (M) Equilibrium (M) A certain weak acid, HA, has a K_a value of 5.5 times 10^-7. Part A Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units. Part B Calculate the percent ionization of HA in a 0.010 M solution. Express your answer to two significant figures, and include the appropriate units.

Explanation / Answer

Part A)

                               HA        <====>       H+         +       A-

Intial                      0.1 M                          -                      -

Change                  -x                             +x                    +x

Equilibrium          0.1-x                           +x                    +x

                              Ka = x2 / 0.1-x = 5.5 * 10-7

x2 = 5.5 * 10-8 - 5.5 * 10-7x

x = 0.0002342 [HA] ionized

% Ionization = (0.0002342 / 0.1) x 100 = 0.23 %

Part B)

                               HA        <====>       H+         +       A-

Intial                      0.01 M                          -                      -

Change                  -x                             +x                    +x

Equilibrium          0.01-x                           +x                    +x

                              Ka = x2 / 0.01-x = 5.5 * 10-7

x2 = 5.5 * 10-9 - 5.5 * 10-8x

x = 0.00007413 [HA] ionized

% Ionization = (0.00007413 / 0.01) x 100 = 0.74%

Part A)

                               HA        <====>       H+         +       A-

Intial                      0.1 M                          -                      -

Change                  -x                             +x                    +x

Equilibrium          0.1-x                           +x                    +x

                              Ka = x2 / 0.1-x = 6.9 * 10-7

x2 = 6.9 * 10-8 - 6.9 * 10-7x

x = 0.000262 [HA] ionized

% Ionization = (0.000262 / 0.1) x 100 = 0.26 %

Part B)

                               HA        <====>       H+         +       A-

Intial                      0.01 M                          -                      -

Change                  -x                             +x                    +x

Equilibrium          0.01-x                           +x                    +x

                              Ka = x2 / 0.01-x = 6.9 * 10-7

x2 = 6.9 * 10-9 - 6.9 * 10-8x

x = 0.000083 [HA] ionized

% Ionization = (0.000083 / 0.01) x 100 = 0.83 %

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