please solve step by step t Percent lonization A certain weak acid, HA, has a Ka

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t Percent lonization A certain weak acid, HA, has a Ka value of 1.5x10- 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: Part A [HA] ionized × 100% Percent ionizationHA initial Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units. 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 Hints A convenient way to keep track of changing concentrations is through what is often called an IC.E table, where l 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 alue Units ahe A B AB Submit My Answers Give Up Initial (M) Change (M) Equilibrium (M) 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. Hints alue Units Submit My Answers Give Up

Explanation / Answer

A)

Lets write the dissociation equation of HA

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

0.1 0 0

0.1-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.5*10^-7)*0.1) = 1.225*10^-4

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

so, x = 1.225*10^-4 M

% dissociation = (x*100)/c

= 1.225*10^-4*100/0.1

= 0.12 %

Answer: 0.12 %

B)

Lets write the dissociation equation of HA

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

1*10^-2 0 0

1*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.5*10^-7)*1*10^-2) = 3.873*10^-5

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

so, x = 3.873*10^-5 M

% dissociation = (x*100)/c

= 3.873*10^-5*100/0.01

= 0.39 %

Answer: 0.39 %

Feel free to comment below if you have any doubts or if this answer do not work

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