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
A 29.9 mL sample of 0.358 M ethylamine, CH3NH2, is titrated with 0.271 M hydroiodic acid....
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