We are asked to determine the volume in mL of 0.3 M HI(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 41 mL of 0.47 M (CH3)2NH(aq).
At the half-equivalence point, half of the moles of (CH3)2NH(aq) will be consumed.
We'll go through the following steps to solve the problem
Step 1. Calculate the moles of (CH3)2NH(aq)
Step 2. Calculate half of the moles of (CH3)2NH(aq)
Step 3. Calculate the volume of 0.3 M needed for half of the moles of (CH3)2NH(aq)
Step 1. Calculate the moles of (CH3)2NH(aq)
recall:
Determine the volume in mL of 0.3 M HI(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 41 mL of 0.47 M (CH3)2NH(aq)(aq). The Kb of dimethylamine is 5.4 x 10-4.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Weak Base Strong Acid Titrations concept. You can view video lessons to learn Weak Base Strong Acid Titrations. Or if you need more Weak Base Strong Acid Titrations practice, you can also practice Weak Base Strong Acid Titrations practice problems.