Question

3. Find the pH of a 2.9M solution of KHSO4 (HSO4 K, = 1.3x 102)

Answer 1

HSO4- dissociates as:

HSO4- -----> H+ + SO42-

1.9 0 0

1.9-x x x

Ka = [H+][SO42-]/[HSO4-]

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 ((1.3*10^-2)*1.9) = 0.1572

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.3*10^-2 = x^2/(1.9-x)

2.47*10^-2 - 1.3*10^-2 *x = x^2

x^2 + 1.3*10^-2 *x-2.47*10^-2 = 0

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

a = 1

b = 1.3*10^-2

c = -2.47*10^-2

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 = 9.897*10^-2

roots are :

x = 0.1508 and x = -0.1638

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

x = 0.1508

So, [H+] = x = 0.1508 M

use:

pH = -log [H+]

= -log (0.1508)

= 0.8216

Answer: 0.822