Question

# Find the pH of the solution obtained when 32 mL of 0.087 M ethylamine, C2H5NH2, is...

 Find the pH of the solution obtained when 32 mL of 0.087 M ethylamine, C2H5NH2, is titrated to the equivalence point with 0.15 M HCl. The value of Kb for ethylamine is 4.7 x 10-4. (in 3 s.f.)

Kb = 4.7 x 10^-4

pKb = 3.328

mmoles of ethylamine = 32 x 0.087 = 2.784

At equivalence point :

mmoles of ethylamine = mmoles of acid

32 x 0.087 = 0.15 x V

V = 18.56 mL

here salt only remains.

salt concentration = 2.784 / 32 + 18.56 = 0.05506 M

pH = 7 - 1/2 (pKa + log C)

= 7 - 1/2 (3.328 + log 0.05506)

pH = 5.97

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