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± Percent Ionization A certain weak acid, HA, has a Ka value of 8.8 10-7 Percent

ID: 545449 • Letter: #

Question

± Percent Ionization A certain weak acid, HA, has a Ka value of 8.8 10-7 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 Calculate the percent ionization of HA in a 0.10 M solution Express your answer to two significant figures and include the appropriate units [HA] ionized × 100% Percent ionization HA initial Hints Percent ionization increases with increasing K. 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. Value Units 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 Tali Submit My Answers Give Up Part B Calculate the percent ionizaton of HA in a 0 010 Msolution. Express your answer to two significant figures, and include the appropriate units A] B IAB Initial (M) Change (M) Equilibrium (M) Hints HA Submit My Answers Give Up

Explanation / Answer

     part-A

   HA(aq) -------------------> H^+ (aq) + A^-(aq)

I           0.1                                  0                  0

C         -x                                  +x                 +x

E           0.1-x                             +x                +x

              ka = [H^+][A^-]/[HA]

              8.8*10^-7   = x*x/(0.1-x)

              8.8*10^-7 *(0.1-x) = x^2

                   x   = 0.000296

       [H^+]   = x   = 0.000296M

     percentage ionisation = [HA] ionisation*100/[HA] initial

                                         = 0.000296*100/0.1   = 0.296%

part-B

            HA(aq) -------------------> H^+ (aq) + A^-(aq)

I           0.1                                  0                  0

C         -x                                  +x                 +x

E           0.1-x                             +x                +x

              ka = [H^+][A^-]/[HA]

              8.8*10^-7   = x*x/(0.1-x)

              8.8*10^-7 *(0.1-x) = x^2

                   x   = 0.000296

       [H^+]   = x   = 0.000296M

     percentage ionisation = [HA] ionisation*100/[HA] initial

                                         = 0.000296*100/0.1   = 0.296%