what is the pH of a 0.035M CH3COOH solution?
CH3COOH dissociates as:
CH3COOH
-----> H+ + CH3COO-
0.035
0 0
0.035-x
x x
Ka = [H+][CH3COO-]/[CH3COOH]
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^-5)*3.5*10^-2) = 7.937*10^-4
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^-5 = x^2/(0.035-x)
6.3*10^-7 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-6.3*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -6.3*10^-7
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 = 2.52*10^-6
roots are :
x = 7.848*10^-4 and x = -8.028*10^-4
since x can't be negative, the possible value of x is
x = 7.848*10^-4
So, [H+] = x = 7.848*10^-4 M
use:
pH = -log [H+]
= -log (7.848*10^-4)
= 3.11
Answer: 3.11
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