Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 25.0 mL of 0.200 M HA with 0.100 M NaOH. Ka = 2.0 x 10-5.
1)when 0.0 mL of NaOH is added
HA dissociates as:
HA -----> H+ + A-
0.2 0 0
0.2-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 ((2*10^-5)*0.2) = 2*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)
2*10^-5 = x^2/(0.2-x)
4*10^-6 - 2*10^-5 *x = x^2
x^2 + 2*10^-5 *x-4*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 2*10^-5
c = -4*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 = 1.6*10^-5
roots are :
x = 1.99*10^-3 and x = -2.01*10^-3
since x can't be negative, the possible value of x is
x = 1.99*10^-3
use:
pH = -log [H+]
= -log (1.99*10^-3)
= 2.7011
Answer: 2.70
2)when 10.0 mL of NaOH is added
Given:
M(HA) = 0.2 M
V(HA) = 25 mL
M(NaOH) = 0.1 M
V(NaOH) = 10 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.2 M * 25 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 10 mL = 1 mmol
We have:
mol(HA) = 5 mmol
mol(NaOH) = 1 mmol
1 mmol of both will react
excess HA remaining = 4 mmol
Volume of Solution = 25 + 10 = 35 mL
[HA] = 4 mmol/35 mL = 0.1143M
[A-] = 1/35 = 0.0286M
They form acidic buffer
acid is HA
conjugate base is A-
Ka = 2*10^-5
pKa = - log (Ka)
= - log(2*10^-5)
= 4.699
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.699+ log {2.857*10^-2/0.1143}
= 4.097
Answer: 4.10
3)when 25.0 mL of NaOH is added
Given:
M(HA) = 0.2 M
V(HA) = 25 mL
M(NaOH) = 0.1 M
V(NaOH) = 25 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.2 M * 25 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol
We have:
mol(HA) = 5 mmol
mol(NaOH) = 2.5 mmol
2.5 mmol of both will react
excess HA remaining = 2.5 mmol
Volume of Solution = 25 + 25 = 50 mL
[HA] = 2.5 mmol/50 mL = 0.05M
[A-] = 2.5/50 = 0.05M
They form acidic buffer
acid is HA
conjugate base is A-
Ka = 2*10^-5
pKa = - log (Ka)
= - log(2*10^-5)
= 4.699
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.699+ log {5*10^-2/5*10^-2}
= 4.699
Answer: 4.70
4)when 50.0 mL of NaOH is added
Given:
M(HA) = 0.2 M
V(HA) = 25 mL
M(NaOH) = 0.1 M
V(NaOH) = 50 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.2 M * 25 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 50 mL = 5 mmol
We have:
mol(HA) = 5 mmol
mol(NaOH) = 5 mmol
5 mmol of both will react to form A- and H2O
A- here is strong base
A- formed = 5 mmol
Volume of Solution = 25 + 50 = 75 mL
Kb of A- = Kw/Ka = 1*10^-14/2*10^-5 = 5*10^-10
concentration ofA-,c = 5 mmol/75 mL = 0.0667M
A- dissociates as
A- + H2O -----> HA + OH-
0.0667 0 0
0.0667-x x x
Kb = [HA][OH-]/[A-]
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 ((5*10^-10)*6.667*10^-2) = 5.774*10^-6
since c is much greater than x, our assumption is correct
so, x = 5.774*10^-6 M
[OH-] = x = 5.774*10^-6 M
use:
pOH = -log [OH-]
= -log (5.774*10^-6)
= 5.2386
use:
PH = 14 - pOH
= 14 - 5.2386
= 8.7614
Answer: 8.76
5)when 60.0 mL of NaOH is added
Given:
M(HA) = 0.2 M
V(HA) = 25 mL
M(NaOH) = 0.1 M
V(NaOH) = 60 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.2 M * 25 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 60 mL = 6 mmol
We have:
mol(HA) = 5 mmol
mol(NaOH) = 6 mmol
5 mmol of both will react
excess NaOH remaining = 1 mmol
Volume of Solution = 25 + 60 = 85 mL
[OH-] = 1 mmol/85 mL = 0.0118 M
use:
pOH = -log [OH-]
= -log (1.176*10^-2)
= 1.9294
use:
PH = 14 - pOH
= 14 - 1.9294
= 12.0706
Answer: 12.07
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