Question

Calculate [H⁺] for a 5.69×10^-2 M aqueous solution of chloroacetic acid (HC₂H₂ClO₂; K_a = 1.4×10^-3).

Calculate the pH of the above solution.

Thanks!

Answer 1

1)
HC2H2ClO2 dissociates as:

HC2H2ClO2 -----> H+ + C2H2ClO2-
5.69*10^-2 0 0
5.69*10^-2-x x x


Ka = [H+][C2H2ClO2-]/[HC2H2ClO2]
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 ((1.4*10^-3)*5.69*10^-2) = 8.925*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)
1.4*10^-3 = x^2/(5.69*10^-2-x)
7.966*10^-5 - 1.4*10^-3 *x = x^2
x^2 + 1.4*10^-3 *x-7.966*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.4*10^-3
c = -7.966*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 = 3.206*10^-4

roots are :
x = 8.253*10^-3 and x = -9.653*10^-3

since x can't be negative, the possible value of x is
x = 8.253*10^-3

so,
[H+] = x = 8.253*10^-3 M

Answer: 8.253*10^-3 M

2)
use:
pH = -log [H+]
= -log (8.253*10^-3)
= 2.083406

Answer: 2.08