Question

The venom of biting ants contains formic acid, HCOOH (Ka = 1.8×10–4 at 25 °C). What is the pH of a 0.105 M solution of formic acid?

Answer 1

HCOOH dissociates as:

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

0.105 0 0

0.105-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)*0.105) = 4.347*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/(0.105-x)

1.89*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-1.89*10^-5 = 0

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

a = 1

b = 1.8*10^-4

c = -1.89*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 = 7.563*10^-5

roots are :

x = 4.258*10^-3 and x = -4.438*10^-3

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

x = 4.258*10^-3

So, [H+] = x = 4.258*10^-3 M

use:

pH = -log [H+]

= -log (4.258*10^-3)

= 2.3708

Answer: 2.37