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**

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

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