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.

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 =

14.3mL you require to reach the half-equivalence point

tallinn tallinn · Answer 2 · 2022-04-02T11:43:15+02:00

When The Kb of ethylamine is 6.5 x 10-4 is = 14.3mL you require to reach the half-equivalence point.

What is Ethylamine?

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

Then CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

So when the reaction is 1:1, to reach the equivalence point then you instruct to that add the moles of HClO₄ equal to the moles CH₃CH₂NH₂ you add originally. Also, When the half-equivalence point requires you to add half-moles of CH₃CH₂NH₂ you add originally.

Then Initial moles of CH₃CH₂NH₂ are:

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

Then 0.0106moles CH₃CH₂NH₂To get the half-equivalence point you require is:

Then 0.0106moles ÷ 2 = 0.005304 moles HClO₄

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

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

Therefore, 14.3mL you require to reach the half-equivalence point.

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