Calculate the pH at the equivalence point for the titration of 0.200 M 0.200 M methylamine ( CH 3 NH 2 ) (CH3NH2) with 0.200 M HCl . 0.200 M HCl. The K b Kb of methylamine is 5.0 × 10 − 4 . 5.0×10−4. pH =

The pH is 10.6

Explanation:

pH = pKa

pKb = - log (5.0 × )

pKb = 3.30

But pKa + pKb = 14 at equivalence point

∴ pKa = 14 - 3.30 = 10.598 ≈ 10.6

⇒pH = 10.6

N.B : What we calculate is not the pH at the end point - but the theoretical pH at the equivalence point.

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