Percent ionization can be used to quantify the extent of ionization of an acid i

Question

Percent ionization can be used to quantify the extent of ionization of an acid in solution and is defined by the following formula for the acid HA:

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 concentrations of reactant(s) and product(s) across the top, creating the columns, and write the rows I.C.E on the left-hand side. Such a table is shown below for the reaction: A+B?AB

Initial (M)Change (M)Equilibrium (M)

A certain weak acid, HA, has a Ka value of 9.9×10?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 dissociates as:

HA -----> H+ + A-

0.1 0 0

0.1-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((9.9*10^-7)*0.1) = 3.146*10^-4

since c is much greater than x, our assumption is correct

so, x = 3.146*10^-4 M

% dissociation = (x*100)/c

= 3.146*10^-4*100/0.1

= 0.315 %

Answer: 0.315 %

B)

HA dissociates as:

HA -----> H+ + A-

1*10^-2 0 0

1*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((9.9*10^-7)*1*10^-2) = 9.95*10^-5

since c is much greater than x, our assumption is correct

so, x = 9.95*10^-5 M

% dissociation = (x*100)/c

= 9.95*10^-5*100/0.01

= 0.995 %

Answer: 0.995 %

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.