CH3NH2 + HCl ____________> CH3NH3Cl
V x0.15 Vx0.15 0 initial mmoles
00 0 Vx0.15 at equivalence
thus the [salt] = mmoles/ total volume = Vx0.15/(V+V) = 0.075 M
pKb of base = -log 5x10-4 = 3.3010
The pH of salt of weak base with strong acid (acidic due to cationic hydrolysis) is given by
pH = 1/2[pKw-pKb - log C]
= 1/2 [ 14 - 3.3010 -log 0.075]
= 5.911
Resources Hint Check Answer < Question 38 of 39 > Calculate the pH at the equivalence...
5.83 is wrong
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH,NH,) with 0.110 M HCI. The Ky of methylamine is 5.0 x 10- pH = 5.83
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Please help
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