quantitative analysis chemistry

Question

How many grams HCl should be added to a solution of 7.00g propylamine(C₃H₇NH₂) (K_a for the conjugate acid =2.72x10^-11) in 1.00L water to produce a buffer of pH 10.4?

Answer 1

C₃H₇NH₂ + H⁺ ==>C₃H₇NH₃⁺

From the Henderson Hasselbalch equation, we can use...

pOH = pKb + log [C₃H₇NH₃⁺ ] / [C₃H₇NH₂ ]

Kb = Kw/Ka = 10^-14 / 2.72x10^-11 = 3.68x10^-4

pKb = -log 3.68x10-4 = 3.43

pOH = 14 - pH = 14 - 10.4 = 3.6

7.00 g C₃H₇NH₂ x 1 mol C₃H₇NH₂ / 59 g = 0.119 mols

HH equation:

3.6 = 3.43 + log [C₃H₇NH₃⁺ ] / [C₃H₇NH₂ ]

log [C₃H₇NH₃⁺ ] / [C₃H₇NH₂ ] = 0.17

[C₃H₇NH₃⁺ ] / [C₃H₇NH₂ ] = 1.48

x / 0.119 - x = 1.48

x = 0.071 mols C₃H₇NH₃⁺ = 0.071 mols HCl

0.071 mols HCl x 36.5 g /mol = 2.59 g HCl