Problem: Determine the volume in mL of 0.23 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 25 mL of 0.48 M CH3CH2NH2(aq)(aq).  The Kb of ethylamine is 6.5 x 10-4.a. 12.55 mLb. 25.10 mLc. 26.09 mLd. 50.20 mLe. 13.05 mL

FREE Expert Solution

We are being asked to determine the volume of the 0.23M HClO4 solution at half-equivalence point upon titration with 25 mL 0.48 M CH3CH2NH(weak base)

Recall that at half-equivalence point, the moles of the weak base will be reduced by half. 

In this case, we have to determine the moles of CH3CH2NH2 using the volume and its molarity.

Then divide it by two and use stoichiometry to relate it to how much HClO4 is needed to reach these moles of weak base

Thus we need to follow the steps:

Step 1. Calculate the moles of CH3CH2NH2 

Step 2. Calculate the moles of HClO4

Step 3. Use dimensional analysis to find V of HClO4

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Problem Details

Determine the volume in mL of 0.23 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 25 mL of 0.48 M CH3CH2NH2(aq)(aq).  The Kb of ethylamine is 6.5 x 10-4.

a. 12.55 mL

b. 25.10 mL

c. 26.09 mL

d. 50.20 mL

e. 13.05 mL

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