use:
pKa = -log Ka
2.87 = -log Ka
Ka = 1.349*10^-3
ClCH2COOH dissociates as:
ClCH2COOH -----> H+ + ClCH2COO-
2.9 0 0
2.9-x x x
Ka = [H+][ClCH2COO-]/[ClCH2COOH]
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 ((1.349*10^-3)*2.9) = 6.255*10^-2
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.349*10^-3 = x^2/(2.9-x)
3.912*10^-3 - 1.349*10^-3 *x = x^2
x^2 + 1.349*10^-3 *x-3.912*10^-3 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.349*10^-3
c = -3.912*10^-3
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 = 1.565*10^-2
roots are :
x = 6.188*10^-2 and x = -6.322*10^-2
since x can't be negative, the possible value of x is
x = 6.188*10^-2
So, [H+] = x = 6.188*10^-2 M
[ClCH2COO-] = x = 6.188*10^-2 M
[ClCH2COOH] = 2.90 - x
= 2.90 - 6.188*10^-2 M
= 2.84 M
use:
pH = -log [H+]
= -log (6.188*10^-2)
= 1.2085
Answer:
[H3O+] = 6.19*10^-2 M
[ClCH2COO-] = 6.19*10^-2 M
[ClCH2COOH] = 2.84 M
pH = 1.21
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