Question

Calculate the pH of a solution that is 0.10M in each of the following: CH₃COOH (K_a = 1.8×10^-5), NaCl, HCl, and HCOOH (K_a = 1.8×10^-4).

Answer 1

1)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.1 0 0

0.1-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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^-5)*0.1) = 1.342*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^-5 = x^2/(0.1-x)

1.8*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

so.[H+] = x = 1.333*10^-3 M

use:

pH = -log [H+]

= -log (1.333*10^-3)

= 2.88

Answer:2.88

b)

NaCl is neutral

so pH will be 7.0

Answer: 7.0

c)

HCl will dissociate completely to give [H+] = 0.10 M

use:

pH = -log [H+]

= -log (0.10)

= 1.0

Answer: 1.0

d)

HCOOH dissociates as:

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

0.1 0 0

0.1-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)*0.1) = 4.243*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/(0.1-x)

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

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

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

a = 1

b = 1.8*10^-4

c = -1.8*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 = 7.203*10^-5

roots are :

x = 4.154*10^-3 and x = -4.334*10^-3

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

x = 4.154*10^-3

so.[H+] = x = 4.154*10^-3 M

use:

pH = -log [H+]

= -log (4.154*10^-3)

= 2.38

Answer: 2.38