We are asked to answer each of the following questions regarding these two titrations.

__Part A__

At the equivalence point, moles of acid = moles of base.

We use the equation below to calculate volume:

$\overline{){\mathbf{\left(}\mathbf{MV}\mathbf{\right)}}_{{\mathbf{a}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{\left(}\mathbf{MV}\mathbf{\right)}}_{{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{M}\mathbf{)}\mathbf{(}\mathbf{20}\mathbf{}\mathbf{mL}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathbf{M}\mathbf{)}\mathbf{\left(}{\mathbf{V}}_{\mathbf{b}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{V}}_{\mathbf{b}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\overline{)\mathbf{M}}\mathbf{)}\mathbf{(}\mathbf{20}\mathbf{}\mathbf{mL}\mathbf{)}\mathbf{}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\overline{)\mathbf{M}}\mathbf{)}}$

**V _{b} = 10 mL KOH**

**10 mL 0.200 M KOH will be needed to titrate 20 mL 0.100 M HCl **

__Part B__

At the equivalence point, moles of acid = moles of base.

We use the equation below to calculate volume:

Two 20.0mL samples, one 0.100 M HCl and the other 0.100 M HF, were titrated with 0.200 M KOH. Answer each of the following questions regarding these two titrations.

Part A

What is the volume of added base at the equivalence point for HCl?

Part B

What is the volume of added base at the equivalence point for HF?

Part C

Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral.

Part D

Predict which titration curve will have the lowest initial pH.