1)
HF dissociates as:
HF -----> H+ + F-
4.5*10^-3 0 0
4.5*10^-3-x x x
Ka = [H+][F-]/[HF]
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 ((6.8*10^-4)*4.5*10^-3) = 1.749*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)
6.8*10^-4 = x^2/(4.5*10^-3-x)
3.06*10^-6 - 6.8*10^-4 *x = x^2
x^2 + 6.8*10^-4 *x-3.06*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.8*10^-4
c = -3.06*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 = 1.27*10^-5
roots are :
x = 1.442*10^-3 and x = -2.122*10^-3
since x can't be negative, the possible value of x is
x = 1.442*10^-3
So, [H+] = x = 1.442*10^-3 M
use:
pH = -log [H+]
= -log (1.442*10^-3)
= 2.841
Answer:
[H+] = 1.44*10^-3 M
pH = 2.84
Only 1 question at a time please
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