Calculate the percent ionization of ascorbic acid
(HC6H7O6) in solutions of each of
the following concentrations (Ka = 8.0e-05.)
(a) 0.185 M
%
(b) 0.575 M
%
(c) 0.740 M
%
a)
HC6H7O6 dissociates as:
HC6H7O6 -----> H+ + C6H7O6-
0.185 0 0
0.185-x x x
Ka = [H+][C6H7O6-]/[HC6H7O6]
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 ((8*10^-5)*0.185) = 3.847*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
8*10^-5 = x^2/(0.185-x)
1.48*10^-5 - 8*10^-5 *x = x^2
x^2 + 8*10^-5 *x-1.48*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8*10^-5
c = -1.48*10^-5
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 5.921*10^-5
roots are :
x = 3.807*10^-3 and x = -3.887*10^-3
since x can't be negative, the possible value of x is
x = 3.807*10^-3
% dissociation = (x*100)/c
= 3.807*10^-3*100/0.185
= 2.058 %
Answer: 2.06 %
B)
HC6H7O6 dissociates as:
HC6H7O6 -----> H+ + C6H7O6-
0.575 0 0
0.575-x x x
Ka = [H+][C6H7O6-]/[HC6H7O6]
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 ((8*10^-5)*0.575) = 6.782*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
8*10^-5 = x^2/(0.575-x)
4.6*10^-5 - 8*10^-5 *x = x^2
x^2 + 8*10^-5 *x-4.6*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8*10^-5
c = -4.6*10^-5
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 1.84*10^-4
roots are :
x = 6.742*10^-3 and x = -6.822*10^-3
since x can't be negative, the possible value of x is
x = 6.742*10^-3
% dissociation = (x*100)/c
= 6.742*10^-3*100/0.575
= 1.173 %
Answer: 1.17 %
C)
HC6H7O6 dissociates as:
HC6H7O6 -----> H+ + C6H7O6-
0.74 0 0
0.74-x x x
Ka = [H+][C6H7O6-]/[HC6H7O6]
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 ((8*10^-5)*0.74) = 7.694*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
8*10^-5 = x^2/(0.74-x)
5.92*10^-5 - 8*10^-5 *x = x^2
x^2 + 8*10^-5 *x-5.92*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8*10^-5
c = -5.92*10^-5
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 2.368*10^-4
roots are :
x = 7.654*10^-3 and x = -7.734*10^-3
since x can't be negative, the possible value of x is
x = 7.654*10^-3
% dissociation = (x*100)/c
= 7.654*10^-3*100/0.74
= 1.034 %
Answer: 1.03 %
Calculate the percent ionization of ascorbic acid (HC6H7O6) in solutions of each of the following concentrations...
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