Question

Calculate the pH of a solution prepared by mixing 20 ml of 9.12 mol/L salicylic acid with 20 ml water

Answer 1

Salicyclic acid is C6H4(OH)COOH.

For simplicity, lets write it as HA

Its Ka is 1.06*10^-3

[HA] = mol of HA / total volume

= M(HA)*V(HA) / total volume

= 9.12 mol/L * 0.020 L / (0.020 L + 0.020 L)

= 4.56 M

HA dissociates as:

HA -----> H+ + A-

4.56 0 0

4.56-x x x

Ka = [H+][A-]/[HA]

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 ((1.06*10^-3)*4.56) = 6.952*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)

1.06*10^-3 = x^2/(4.56-x)

4.834*10^-3 - 1.06*10^-3 *x = x^2

x^2 + 1.06*10^-3 *x-4.834*10^-3 = 0

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

a = 1

b = 1.06*10^-3

c = -4.834*10^-3

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 = 1.934*10^-2

roots are :

x = 6.9*10^-2 and x = -7.006*10^-2

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

x = 6.9*10^-2

So, [H+] = x = 6.9*10^-2 M

use:

pH = -log [H+]

= -log (6.9*10^-2)

= 1.1612

Answer: 1.16