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 adds, 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 9.5 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

A)
HA <----> H+ + A-
0.10 0 0 (initial)
0.10-x x x (at equilibrium)

Ka = x*x/0.10
9.5*10^-7 = x^2 / (0.10)
x = 3.08*10^-4 M

percent ionization = x*100/C
= (3.08*10^-4)*100 / 0.10
=0.31 %

B)
HA <----> H+ + A-
0.010 0 0 (initial)
0.010-x x x (at equilibrium)

Ka = x*x/0.10
9.5*10^-7 = x^2 / (0.010)
x = 9.75*10^-5 M

percent ionization = x*100/C
= (9.75*10^-5)*100 / 0.010
=0.97 %

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