Question

23.

A)Calculate the pH of a 0.475 M aqueous solution of hypochlorous acid (HClO, K_a = 3.5×10^-8).
pH =

B)Calculate the pH of a 0.0354 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).
pH =

Answer 1

A)

HClO dissociates as:

HClO -----> H+ + ClO-

0.475 0 0

0.475-x x x

Ka = [H+][ClO-]/[HClO]

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 ((3.5*10^-8)*0.475) = 1.289*10^-4

since c is much greater than x, our assumption is correct

so, x = 1.289*10^-4 M

So, [H+] = x = 1.289*10^-4 M

use:

pH = -log [H+]

= -log (1.289*10^-4)

= 3.8896

Answer: 3.89

2)

HF dissociates as:

HF -----> H+ + F-

3.54*10^-2 0 0

3.54*10^-2-x x x

Ka = [H+][F-]/[HF]

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 ((7.2*10^-4)*3.54*10^-2) = 5.049*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)

7.2*10^-4 = x^2/(3.54*10^-2-x)

2.549*10^-5 - 7.2*10^-4 *x = x^2

x^2 + 7.2*10^-4 *x-2.549*10^-5 = 0

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

a = 1

b = 7.2*10^-4

c = -2.549*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 = 1.025*10^-4

roots are :

x = 4.701*10^-3 and x = -5.421*10^-3

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

x = 4.701*10^-3

So, [H+] = x = 4.701*10^-3 M

use:

pH = -log [H+]

= -log (4.701*10^-3)

= 2.3278

Answer: 2.33