Question
Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka =...
Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka = 4.0×10-10) by 0.429 M NaOH at the following points.
(a) Before the addition of any NaOH _______
(b) After the addition of 5.30 mL of NaOH _________
(c) At the half-equivalence point (the titration midpoint) ________
(d) At the equivalence point ________
(e) After the addition of 30.7 mL of NaOH ________
Answers
a)when 0.0 mL of NaOH is added
HCN dissociates as:
HCN -----> H+ + CN-
0.406 0 0
0.406-x x x
Ka = [H+][CN-]/[HCN]
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 ((4*10^-10)*0.406) = 1.274*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.274*10^-5 M
use:
pH = -log [H+]
= -log (1.274*10^-5)
= 4.8947
Answer: 4.89
b)when 5.3 mL of NaOH is added
Given:
M(HCN) = 0.406 M
V(HCN) = 21.6 mL
M(NaOH) = 0.429 M
V(NaOH) = 5.3 mL
mol(HCN) = M(HCN) * V(HCN)
mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.429 M * 5.3 mL = 2.2737 mmol
We have:
mol(HCN) = 8.7696 mmol
mol(NaOH) = 2.2737 mmol
2.2737 mmol of both will react
excess HCN remaining = 6.4959 mmol
Volume of Solution = 21.6 + 5.3 = 26.9 mL
[HCN] = 6.4959 mmol/26.9 mL = 0.2415M
[CN-] = 2.2737/26.9 = 0.0845M
They form acidic buffer
acid is HCN
conjugate base is CN-
Ka = 4*10^-10
pKa = - log (Ka)
= - log(4*10^-10)
= 9.398
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.398+ log {8.452*10^-2/0.2415}
= 8.942
Answer: 8.94
c)
At half equivalence point:
pH = pKa
= 9.398
Answer: 9.40
d)
find the volume of NaOH used to reach equivalence point
M(HCN)*V(HCN) =M(NaOH)*V(NaOH)
0.406 M *21.6 mL = 0.429M *V(NaOH)
V(NaOH) = 20.442 mL
Given:
M(HCN) = 0.406 M
V(HCN) = 21.6 mL
M(NaOH) = 0.429 M
V(NaOH) = 20.442 mL
mol(HCN) = M(HCN) * V(HCN)
mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.429 M * 20.442 mL = 8.7696 mmol
We have:
mol(HCN) = 8.7696 mmol
mol(NaOH) = 8.7696 mmol
8.7696 mmol of both will react to form CN- and H2O
CN- here is strong base
CN- formed = 8.7696 mmol
Volume of Solution = 21.6 + 20.442 = 42.042 mL
Kb of CN- = Kw/Ka = 1*10^-14/4*10^-10 = 2.5*10^-5
concentration ofCN-,c = 8.7696 mmol/42.042 mL = 0.2086M
CN- dissociates as
CN- + H2O -----> HCN + OH-
0.2086 0 0
0.2086-x x x
Kb = [HCN][OH-]/[CN-]
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 ((2.5*10^-5)*0.2086) = 2.284*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)
2.5*10^-5 = x^2/(0.2086-x)
5.215*10^-6 - 2.5*10^-5 *x = x^2
x^2 + 2.5*10^-5 *x-5.215*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 2.5*10^-5
c = -5.215*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 = 2.086*10^-5
roots are :
x = 2.271*10^-3 and x = -2.296*10^-3
since x can't be negative, the possible value of x is
x = 2.271*10^-3
[OH-] = x = 2.271*10^-3 M
use:
pOH = -log [OH-]
= -log (2.271*10^-3)
= 2.6438
use:
PH = 14 - pOH
= 14 - 2.6438
= 11.3562
Answer: 11.36
e)when 30.7 mL of NaOH is added
Given:
M(HCN) = 0.406 M
V(HCN) = 21.6 mL
M(NaOH) = 0.429 M
V(NaOH) = 30.7 mL
mol(HCN) = M(HCN) * V(HCN)
mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.429 M * 30.7 mL = 13.1703 mmol
We have:
mol(HCN) = 8.7696 mmol
mol(NaOH) = 13.1703 mmol
8.7696 mmol of both will react
excess NaOH remaining = 4.4007 mmol
Volume of Solution = 21.6 + 30.7 = 52.3 mL
[OH-] = 4.4007 mmol/52.3 mL = 0.0841 M
use:
pOH = -log [OH-]
= -log (8.414*10^-2)
= 1.075
use:
PH = 14 - pOH
= 14 - 1.075
= 12.925
Answer: 12.93