Question

1. Calculate the pH of a 0.434 M aqueous solution of acetylsalicylic acid (aspirin)(HC₉H₇O₄, K_a = 3.0×10^-4).

2. Calculate the pH of a 0.0420 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).

3. Calculate the pH of a 0.0973 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).

Answer 1

1)

HC9H7O4 dissociates as:

HC9H7O4 -----> H+ + C9H7O4-

0.434 0 0

0.434-x x x

Ka = [H+][C9H7O4-]/[HC9H7O4]

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*10^-4)*0.434) = 1.141*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)

3*10^-4 = x^2/(0.434-x)

1.302*10^-4 - 3*10^-4 *x = x^2

x^2 + 3*10^-4 *x-1.302*10^-4 = 0

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

a = 1

b = 3*10^-4

c = -1.302*10^-4

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 = 5.209*10^-4

roots are :

x = 1.126*10^-2 and x = -1.156*10^-2

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

x = 1.126*10^-2

So, [H+] = x = 1.126*10^-2 M

use:

pH = -log [H+]

= -log (1.126*10^-2)

= 1.9484

Answer: 1.95

2)

HF dissociates as:

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

4.2*10^-2 0 0

4.2*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)*4.2*10^-2) = 5.499*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/(4.2*10^-2-x)

3.024*10^-5 - 7.2*10^-4 *x = x^2

x^2 + 7.2*10^-4 *x-3.024*10^-5 = 0

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

a = 1

b = 7.2*10^-4

c = -3.024*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 = 1.215*10^-4

roots are :

x = 5.151*10^-3 and x = -5.871*10^-3

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

x = 5.151*10^-3

So, [H+] = x = 5.151*10^-3 M

use:

pH = -log [H+]

= -log (5.151*10^-3)

= 2.2881

Answer: 2.29

3)

HF dissociates as:

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

9.73*10^-2 0 0

9.73*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)*9.73*10^-2) = 8.37*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/(9.73*10^-2-x)

7.006*10^-5 - 7.2*10^-4 *x = x^2

x^2 + 7.2*10^-4 *x-7.006*10^-5 = 0

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

a = 1

b = 7.2*10^-4

c = -7.006*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 = 2.807*10^-4

roots are :

x = 8.018*10^-3 and x = -8.738*10^-3

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

x = 8.018*10^-3

So, [H+] = x = 8.018*10^-3 M

use:

pH = -log [H+]

= -log (8.018*10^-3)

= 2.096

Answer: 2.10