Question

What is the pH of a 0.01 M solution of formic acid ? The ka of formic acid is 1.77 x 10-4

Answer 1

HCOOH dissociates as:

HCOOH          ----->     H+   + HCOO-
1*10^-2                 0         0
1*10^-2-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 ((1.77*10^-4)*1*10^-2) = 1.33*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)
1.77*10^-4 = x^2/(1*10^-2-x)
1.77*10^-6 - 1.77*10^-4 *x = x^2
x^2 + 1.77*10^-4 *x-1.77*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.77*10^-4
c = -1.77*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 = 7.111*10^-6

roots are :
x = 1.245*10^-3 and x = -1.422*10^-3

since x can't be negative, the possible value of x is
x = 1.245*10^-3

So, [H+] = x = 1.245*10^-3 M


use:
pH = -log [H+]
= -log (1.245*10^-3)
= 2.90
Answer: 2.90