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?
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
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