Question

Enter your answer in the provided box. Calculate the pH at the equivalence point in the titration of 45.0 mL of 0.115 M methylamine (Kb = 4.4 x 10-4) with 0.345 M HCI. pH = 1

Answer 1

find the volume of HCl used to reach equivalence point

M(CH3NH2)*V(CH3NH2) =M(HCl)*V(HCl)

0.115 M *45.0 mL = 0.345M *V(HCl)

V(HCl) = 15 mL

Given:

M(HCl) = 0.345 M

V(HCl) = 15 mL

M(CH3NH2) = 0.115 M

V(CH3NH2) = 45 mL

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

mol(HCl) = 0.345 M * 15 mL = 5.175 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.115 M * 45 mL = 5.175 mmol

We have:

mol(HCl) = 5.175 mmol

mol(CH3NH2) = 5.175 mmol

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

CH3NH3+ here is strong acid

CH3NH3+ formed = 5.175 mmol

Volume of Solution = 15 + 45 = 60 mL

Ka of CH3NH3+ = Kw/Kb = 1.0E-14/4.4E-4 = 2.273*10^-11

concentration ofCH3NH3+,c = 5.175 mmol/60 mL = 0.0862 M

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

8.625*10^-2 0 0

8.625*10^-2-x x x

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

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 ((2.273*10^-11)*8.625*10^-2) = 1.4*10^-6

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

so, x = 1.4*10^-6 M

[H+] = x = 1.4*10^-6 M

use:

pH = -log [H+]

= -log (1.4*10^-6)

= 5.8538

Answer: 5.85