Calculate the pH at 25°C of a 0.075 M solution of a weak acid that has Ka = 1.3 ×10−5.
Let the weak acid be written as HA
HA dissociates as:
HA -----> H+ + A-
7.5*10^-2 0 0
7.5*10^-2-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 ((1.3*10^-5)*7.5*10^-2) = 9.874*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)
1.3*10^-5 = x^2/(7.5*10^-2-x)
9.75*10^-7 - 1.3*10^-5 *x = x^2
x^2 + 1.3*10^-5 *x-9.75*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.3*10^-5
c = -9.75*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 = 3.9*10^-6
roots are :
x = 9.809*10^-4 and x = -9.939*10^-4
since x can't be negative, the possible value of x is
x = 9.809*10^-4
so.[H+] = x = 9.809*10^-4 M
use:
pH = -log [H+]
= -log (9.809*10^-4)
= 3.01
Answer: 3.01
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