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**Problem**: Determine the volume in mL of 0.33 M Hl(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 25 mL of 0.24 M (CH3)2NH(aq)(aq). The Kb of dimethylamine is 5.4 x 10-4. a. 2.27 mLb. 4.55 mLc. 9.09 mLd. 14.2 mLe. 28.4 mL

###### FREE Expert Solution

We are being asked to determine the **volume of the 0.33 M HI solution** at half-equivalence point upon **titration with 25 mL 0.24 M (CH _{3})_{2}NH**

_{ }

**(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 (CH_{3})_{2}NH using the volume and its molarity.

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

Thus we need to follow the steps:

**Step 1. **Calculate the moles of (CH_{3})_{2}NH

**Step 2. **Calculate the moles of HI

**Step 3. **Use dimensional analysis to find V of HI

###### Problem Details

Determine the volume in mL of 0.33 M Hl(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 25 mL of 0.24 M (CH_{3})_{2}NH(aq)(aq). The K_{b} of dimethylamine is 5.4 x 10^{-4}.

a. 2.27 mL

b. 4.55 mL

c. 9.09 mL

d. 14.2 mL

e. 28.4 mL

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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 .

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