Question

Determine the following in a 0.0390 M solution of HIO₃ (K_a = 0.170)?

[HIO3] =	_____ M
[IO31-] =	_____M
[H3O1+] =	_____M
pH =	_____

Answer 1

HIO3 dissociates as:

HIO3 -----> H+ + IO3-

3.9*10^-2 0 0

3.9*10^-2-x x x

Ka = [H+][IO3-]/[HIO3]

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 ((0.17)*3.9*10^-2) = 8.142*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)

0.17 = x^2/(3.9*10^-2-x)

6.63*10^-3 - 0.17 *x = x^2

x^2 + 0.17 *x-6.63*10^-3 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 0.17

c = -6.63*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 = 5.542*10^-2

roots are :

x = 3.271*10^-2 and x = -0.2027

since x can't be negative, the possible value of x is

x = 3.271*10^-2

So, [H+] = x = 3.271*10^-2 M

[IO3-] = x = 3.271*10^-2 M

[HIO3] = 0.0390 - x

= 0.0390 - 3.271*10^-2

= 6.29*10^-3 M

use:

pH = -log [H+]

= -log (3.271*10^-2)

= 1.4854

[HIO3] =6.29*10^-3 M

[IO3-] = 3.271*10^-2 M

[H3O+] = 3.27*10^-2 M

pH = 1.49