Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 20.8 mL of 0.51 M CH3CH2NH2(aq). Enter your answer with one decimal place. The Kb of ethylamine is 6.5 x 10-4.

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

Initial moles of CH₃CH₂NH₂ are:

20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

0.0106moles CH₃CH₂NH₂

To reach the half-equivalence point you require:

0.0106moles ÷ 2 = 0.005304 moles HClO₄

As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error