Question

22.

A)The pH of an aqueous solution of 0.475 M hypochlorous acid is

B)The pOH of an aqueous solution of 0.475 M formic acid, HCOOH is

Answer 1

A)

Ka of HClO = 3.5*10^-8

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)

Ka of HCOOH = 3.5*10^-8

HCOOH dissociates as:

HCOOH -----> H+ + HCOO-

0.475 0 0

0.475-x x x

Ka = [H+][HCOO-]/[HCOOH]

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 ((1.8*10^-4)*0.475) = 9.247*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)

1.8*10^-4 = x^2/(0.475-x)

8.55*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-8.55*10^-5 = 0

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

a = 1

b = 1.8*10^-4

c = -8.55*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 = 3.42*10^-4

roots are :

x = 9.157*10^-3 and x = -9.337*10^-3

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

x = 9.157*10^-3

So, [H+] = x = 9.157*10^-3 M

use:

pH = -log [H+]

= -log (9.157*10^-3)

= 2.04

Now use:

pOH = 14 - pH

= 14 - 2.04

= 11.96

Answer: 11.96