Question

Calculate the pH at the equivalence point for the titration of 0.251 M B: (a weak base with pkb = 5.27) with 0.251 M HCI.

Answer 1

use:

pKb = -log Kb

5.27= -log Kb

Kb = 5.37*10^-6

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) = 0.251 M

V(HCl) = 1 mL

M(B) = 0.251 M

V(B) = 1 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.251 M * 1 mL = 0.251 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.251 M * 1 mL = 0.251 mmol

We have:

mol(HCl) = 0.251 mmol

mol(B) = 0.251 mmol

0.251 mmol of both will react to form BH+ and H2O

BH+ here is strong acid

BH+ formed = 0.251 mmol

Volume of Solution = 1 + 1 = 2 mL

Ka of BH+ = Kw/Kb = 1.0E-14/5.370317963702533E-6 = 1.862*10^-9

concentration ofBH+,c = 0.251 mmol/2 mL = 0.1255 M

BH+ + H2O -----> B + H+

0.1255 0 0

0.1255-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 ((1.862*10^-9)*0.1255) = 1.529*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.529*10^-5 M

[H+] = x = 1.529*10^-5 M

use:

pH = -log [H+]

= -log (1.529*10^-5)

= 4.8157

Answer: 4.82