Question

Calculate the pH at the equivalence point for the titration of 0.614 M B: (a weak base with pKb = 4.13) with 0.614 M HCl.

Answer 1

Given:
pKb = 4.13

use:
pKb = -log Kb
4.13= -log Kb
Kb = 7.413*10^-5
Since concentration of both acid and base are same. The volume required for both to reach equivalence point will be same
Let 1 mL of both acid and base are required
Given:
M(HCl) = 0.614 M
V(HCl) = 1 mL
M(B) = 0.614 M
V(B) = 1 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.614 M * 1 mL = 0.614 mmol

mol(B) = M(B) * V(B)
mol(B) = 0.614 M * 1 mL = 0.614 mmol



We have:
mol(HCl) = 0.614 mmol
mol(B) = 0.614 mmol

0.614 mmol of both will react to form BH+ and H2O
BH+ here is acid
BH+ formed = 0.614 mmol
Volume of Solution = 1 + 1 = 2 mL
Ka of BH+ = Kw/Kb = 1.0E-14/7.413102413009176E-5 = 1.349*10^-10
concentration ofBH+,c = 0.614 mmol/2 mL = 0.307 M


BH+      + H2O ----->     B   +   H+
0.307                    0         0
0.307-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 ((1.349*10^-10)*0.307) = 6.435*10^-6

since c is much greater than x, our assumption is correct
so, x = 6.435*10^-6 M



[H+] = x = 6.435*10^-6 M

use:
pH = -log [H+]
= -log (6.435*10^-6)
= 5.1914
Answer: 5.19