Grade 12th PassPhysical Chemistry
Plz ans. the question which is in attachment. Plzzzzzzzzzzzzz.
1 Answer
Arun
Dear Parth
After the reaction: 0.02 mol CH3NH2 and 0.08 mol CH3NH3
by Henderson-Hasselbalch Equation:
pH = pKa + log (base/acid)
pKb = - log 5 x 10?4
= 3.30103
= 3.30103
pKa = 14 - 3.30103
= 10.69897
= 10.69897
pH = 10.69897 + log (0.02 / 0.08)
pH = 10.69897 + (-0.60206)
pH = 10.09691
Antilog the pH to get the H+ conc. which will be = 8 x 10?¹¹ M
Regards
Arun (askIITians forum expert)
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