Question

± 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+ B→AB

A certain weak acid, HA, has a Ka value of 8.5×10−7. 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. 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

Answer 1

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%