Question

A 29.9 mL sample of 0.358 M ethylamine, CH3NH2, is titrated with 0.271 M hydroiodic acid. After adding 17.0 mL of hydroiodic acid, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

C2H5NH2 + HI -------------> CH5NH3⁺ + I^-

initially

millimoles of base = 29.9 x 0.358 = 10.7042

millimoles of acid added = 17.0 x 0.271 = 4.607

after HI added

millimoles of salt formed = 4.607

millimoles of base left = 10.7042 - 4.607 = 6.0972

total volume = 17.0 + 29.9 = 46.9 mL

[salt] = 4.607 / 46.9 = 0.098 M

[base] = 6.0972 / 46.9 = 0.130 M

mixture of weak base and its salt act as basic buffer

for such buffer

pOH = pKb + log [salt] / [base ]

pKb of C2H5NH2 = 3.36

pOH = 3.36 + log [0.098] / [0.130]

pOH = 3.24

pH = 14 - 3.24

pH = 10.76