Question

In the titration of a 25 mL of 0.245 M weak base (Kb = 1.76*10^-5) being titrated by 0.365 M HCl determine the following:

a. The PH at the initial point

b. The PH after 12.3 mL of HCl has been added

c. The PH at the equivalence point

d. The PH after 18.4 mL of HCl has been added

Answer 1

a)when 0.0 mL of HCl is added
B dissociates as:

B +H2O     ----->     BH+   +   OH-
0.245                   0         0
0.245-x                 x         x


Kb = [BH+][OH-]/[B]
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 ((1.76*10^-5)*0.245) = 2.077*10^-3

since c is much greater than x, our assumption is correct
so, x = 2.077*10^-3 M



So, [OH-] = x = 2.077*10^-3 M


use:
pOH = -log [OH-]
= -log (2.077*10^-3)
= 2.6827


use:
PH = 14 - pOH
= 14 - 2.6827
= 11.3173
Answer: 11.32

b)when 12.3 mL of HCl is added
Given:
M(HCl) = 0.365 M
V(HCl) = 12.3 mL
M(B) = 0.245 M
V(B) = 25 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.365 M * 12.3 mL = 4.4895 mmol

mol(B) = M(B) * V(B)
mol(B) = 0.245 M * 25 mL = 6.125 mmol



We have:
mol(HCl) = 4.4895 mmol
mol(B) = 6.125 mmol

4.4895 mmol of both will react
excess B remaining = 1.6355 mmol
Volume of Solution = 12.3 + 25 = 37.3 mL
[B] = 1.6355 mmol/37.3 mL = 0.0438 M
[BH+] = 4.4895 mmol/37.3 mL = 0.1204 M

They form basic buffer
base is B
conjugate acid is BH+

Kb = 1.76*10^-5

pKb = - log (Kb)
= - log(1.76*10^-5)
= 4.754

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.754+ log {0.1204/4.385*10^-2}
= 5.193

use:
PH = 14 - pOH
= 14 - 5.193
= 8.807

Answer: 8.81

c)
find the volume of HCl used to reach equivalence point
M(B)*V(B) =M(HCl)*V(HCl)
0.245 M *25.0 mL = 0.365M *V(HCl)
V(HCl) = 16.7808 mL
Given:
M(HCl) = 0.365 M
V(HCl) = 16.7808 mL
M(B) = 0.245 M
V(B) = 25 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.365 M * 16.7808 mL = 6.125 mmol

mol(B) = M(B) * V(B)
mol(B) = 0.245 M * 25 mL = 6.125 mmol



We have:
mol(HCl) = 6.125 mmol
mol(B) = 6.125 mmol

6.125 mmol of both will react to form BH+ and H2O
BH+ here is acid
BH+ formed = 6.125 mmol
Volume of Solution = 16.7808 + 25 = 41.7808 mL

use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/1.76*10^-5
Ka = 5.682*10^-10
concentration ofBH+,c = 6.125 mmol/41.7808 mL = 0.1466 M


BH+      + H2O ----->     B   +   H+
0.1466                    0         0
0.1466-x                  x         x


Ka = [H+][B]/[BH+]
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 ((5.682*10^-10)*0.1466) = 9.127*10^-6

since c is much greater than x, our assumption is correct
so, x = 9.127*10^-6 M



[H+] = x = 9.127*10^-6 M

use:
pH = -log [H+]
= -log (9.127*10^-6)
= 5.0397
Answer: 5.04

d)when 18.4 mL of HCl is added
Given:
M(HCl) = 0.365 M
V(HCl) = 18.4 mL
M(B) = 0.245 M
V(B) = 25 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.365 M * 18.4 mL = 6.716 mmol

mol(B) = M(B) * V(B)
mol(B) = 0.245 M * 25 mL = 6.125 mmol



We have:
mol(HCl) = 6.716 mmol
mol(B) = 6.125 mmol

6.125 mmol of both will react
excess HCl remaining = 0.591 mmol
Volume of Solution = 18.4 + 25 = 43.4 mL
[H+] = 0.591 mmol/43.4 mL = 0.0136 M

use:
pH = -log [H+]
= -log (1.362*10^-2)
= 1.8659
Answer: 1.87