Question

Determine the pH of a 0.046 M iodous acid (HIO2) solution. Iodous acid is a weak acid (Ka = 3.2 ✕ 10−5 M).

Answer 1

HIO2 dissociates as:

HIO2          ----->     H+   + IO2-
4.6*10^-2                 0         0
4.6*10^-2-x               x         x


Ka = [H+][IO2-]/[HIO2]
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 ((3.2*10^-5)*4.6*10^-2) = 1.213*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)
3.2*10^-5 = x^2/(4.6*10^-2-x)
1.472*10^-6 - 3.2*10^-5 *x = x^2
x^2 + 3.2*10^-5 *x-1.472*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 3.2*10^-5
c = -1.472*10^-6

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 = 5.889*10^-6

roots are :
x = 1.197*10^-3 and x = -1.229*10^-3

since x can't be negative, the possible value of x is
x = 1.197*10^-3

So, [H+] = x = 1.197*10^-3 M


use:
pH = -log [H+]
= -log (1.197*10^-3)
= 2.9218
Answer: 2.92