Question

Calculate the pH at the equivalence point for the titration of 0.250 M0.250 M methylamine (CH3NH2)(CH3NH2) with 0.250 M HCl.0.250 M HCl. The ?bKb of methylamine is 5.0×10−4.5.0×10−4.

pH=pH=

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
Given:
M(HCl) = 0.25 M
V(HCl) = 1 mL
M(CH3NH2) = 0.25 M
V(CH3NH2) = 1 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.25 M * 1 mL = 0.25 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.25 M * 1 mL = 0.25 mmol



We have:
mol(HCl) = 0.25 mmol
mol(CH3NH2) = 0.25 mmol

0.25 mmol of both will react to form CH3NH3+ and H2O
CH3NH3+ here is strong acid
CH3NH3+ formed = 0.25 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 = 0.25 mmol/2 mL = 0.125 M


CH3NH3+      + H2O ----->     CH3NH2   +   H+
0.125                    0         0
0.125-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*10^-11)*0.125) = 1.581*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.581*10^-6 M



[H+] = x = 1.581*10^-6 M

use:
pH = -log [H+]
= -log (1.581*10^-6)
= 5.801
Answer: 5.80