Question

Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer 1

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

we have:

Molarity of HCl = 0.18 M

Volume of HCl = 1 mL

Molarity of CH3NH2 = 0.18 M

Volume of CH3NH2 = 1 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.18 M * 1 mL = 0.18 mmol

mol of CH3NH2 = Molarity of CH3NH2 * Volume of CH3NH2

mol of CH3NH2 = 0.18 M * 1 mL = 0.18 mmol

We have:

mol of HCl = 0.18 mmol

mol of CH3NH2 = 0.18 mmol

0.18 mmol of both will react to form CH3NH3+ and H2O

CH3NH3+ here is strong acid

CH3NH3+ formed = 0.18 mmol

Volume of Solution = 1 + 1 = 2 mL

Ka of CH3NH3+ = Kw/Kb = 1.0E-14/5.0E-4 = 2*10^-11

concentration ofCH3NH3+,c = 0.18 mmol/2 mL = 0.09 M

CH3NH3+ + H2O -----> CH3NH2 + H+

9*10^-2 0 0

9*10^-2-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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^-11)*9*10^-2) = 1.342*10^-6

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

so, x = 1.342*10^-6 M

[H+] = x = 1.342*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (1.342*10^-6)

= 5.87

Answer: 5.87