How do you calculate the volume?

Question

Determine the volume in mL of 0.53 M HClO₄(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 41.3 mL of 0.55 M CH₃CH₂NH₂(aq).

Answer 1

CH₃CH₂NH₂ + H+ ==> CH₃CH₂NH₃⁺

Initial moles CH₃CH₂NH₂ = 0.0413 L x 0.55 mol/L = 0.0227 moles

Initial moles CH₃CH₂NH₃⁺ = 0

At the half way point, 1/2 of the CH₃CH₂NH₂ has been converted to CH₃CH₂NH₃⁺

At the half way point you have 0.01135 moles CH₃CH₂NH₂ and 0.01135 moles CH₃CH₂NH₃⁺

This requires 0.01135 moles of HCl

Volume of HCl: (x L)(0.53 mol/L) = 0.01135 moles

x = 0.0214 L = 21.4 mls HCl needed