In addition to the acid-dissociation constant, K a, another measure of the stren

Question

In addition to the acid-dissociation constant, Ka, another measure of the strength of an acid is percent ionization, determined by the following formula:

Percent ionization=[HA] ionized[HA] initial×100%

Percent ionization increases with increasing Ka. Strong acids, for which Ka 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.

Initial (M)Change (M)Equilibrium (M)A+ BAB

A certain weak acid, HA, has a Ka value of 3.9×107.

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 :

Let a be the dissociation of the weak acid
                            HA <---> H + + A-

initial conc.            c               0         0

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , Ka = ca x ca / ( c(1-a)

                                         = c a2 / (1-a)

In the case of weak acids is very small so 1-a is taken as 1

So Ka = ca2

==> a = ( Ka / c )

Given Ka = 3.9x10-7

          c = concentration = 0.10 M

Plug the values we get a = ( Ka / c )

                                        = ((3.9x10-7)/ 0.10)

                                        = 1.97 x10-3

% dissociation = 1.97x10-3 x 100 = 0.197

Part B :

Let a be the dissociation of the weak acid
                            HA <---> H + + A-

initial conc.            c               0         0

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , Ka = ca x ca / ( c(1-a)

                                         = c a2 / (1-a)

In the case of weak acids is very small so 1-a is taken as 1

So Ka = ca2

==> a = ( Ka / c )

Given Ka = 3.9x10-7

          c = concentration = 0.010 M

Plug the values we get a = ( Ka / c )

                                        = ((3.9x10-7)/ 0.010)

                                        = 6.24 x10-3

% dissociation = 6.24x10-3 x 100 = 0.62

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