± Percent Ionization Percent ionization=[HA] ionized[HA] initial×100% Percent io
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± Percent Ionization
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 8.5×107.
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.
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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.
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Provi
± Percent Ionization
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 8.5×107.
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.
SubmitHintsMy AnswersGive UpReview Part
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.
SubmitHintsMy AnswersGive UpReview Part
Provi
Explanation / Answer
1. HA -----> H+ + A-
I 0.10 0 0
C -x +x +x
E 0.10-x x x
Ka = x.x/0.10-x
8.5*10^-7 = x2/0.10-x
value of x is smaller than 0.10, so 0.10-x can be written as 0.10
8.5*10^-7 = x2/0.10
x2= 8.5*10^-7. 0.10= 85*10^-10
x= 9.21*10^-5
% ionization= [HA] ionized/[HA]initial*100 = 9.21*10^-5/0.10* 100= 0.92%
2. HA -----> H+ + A-
I 0.010 0 0
C -x +x +x
E 0.010-x x x
Ka= x.x/0.010-x
8.5*10^-7 = x2/0.010
x2= 8.5*10^-7*0.010= 8.5*10^-10
x= 2.91*10^-5
% ionization = 2.91*10^-5/0.010*100= 2.91%
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