calculate the pH and pOH of ca
calculate the pH and pOH of cach of the following soulutions at25 degrees C
a. 0.15 M HNF [Ka(HF) = 3.5 x 10-4]
b. 0.15 m NH3 [Kb (NH3) = 1.8 x10-5]
Answer:
a)
HF dissociates as:
HF —–> H+ + F-
0.15 0 0
0.15-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 ((3.5*10^-4)*0.15) = 7.246*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)
3.5*10^-4 = x^2/(0.15-x)
5.25*10^-5 – 3.5*10^-4 *x = x^2
x^2 + 3.5*10^-4 *x-5.25*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 3.5*10^-4
c = -5.25*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 = 2.101*10^-4
roots are :
x = 7.073*10^-3 and x = -7.423*10^-3
since x can’t be negative, the possible value of x is
x = 7.073*10^-3
So, [H+] = x = 7.073*10^-3 M
use:
pH = -log [H+]
= -log (7.073*10^-3)
= 2.1504
pOH = 14 – pH
= 14 – 2.15
= 11.85
pH = 2.15
pOH = 11.85
b)
NH3 dissociates as:
NH3 +H2O —–> NH4+ + OH-
0.15 0 0
0.15-x x x
Kb = [NH4+][OH-]/[NH3]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.8*10^-5)*0.15) = 1.643*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
1.8*10^-5 = x^2/(0.15-x)
2.7*10^-6 – 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-2.7*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -2.7*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 = 1.08*10^-5
roots are :
x = 1.634*10^-3 and x = -1.652*10^-3
since x can’t be negative, the possible value of x is
x = 1.634*10^-3
So, [OH-] = x = 1.634*10^-3 M
use:
pOH = -log [OH-]
= -log (1.634*10^-3)
= 2.7867
use:
PH = 14 – pOH
= 14 – 2.7867
= 11.2133
pH = 11.21
pOH = 2.79
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