**Calculate the volume needed to reach the equivalence point:**

${\mathbf{M}}_{\mathbf{A}}{\mathbf{V}}_{\mathbf{A}}\mathbf{=}{\mathbf{M}}_{\mathbf{B}}{\mathbf{V}}_{\mathbf{B}}\phantom{\rule{0ex}{0ex}}\frac{(50.0\mathrm{mL})\left(\overline{)1.5M}\right)}{\overline{)\mathbf{1}\mathbf{.}\mathbf{}\mathbf{5}\mathbf{}\mathbf{M}}}\mathbf{=}\frac{\overline{)(1.5M)}\left({V}_{B}\right)}{\overline{)\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{M}}}$

**V _{B} = 50.0 mL**

**To reach the 2nd equivalence point, we will need another 50.00 mL or a total of 100 mL of the base. Therefore, we reached the 2nd equivalence point.**

**After the first equivalence point, we will have H _{2}PO_{3}^{-}. Write the chemical equation for the reaction between H_{2}PO_{3}^{-} and NaOH.**

**Reaction:**

**H _{2}PO_{3}^{-}**

*

**Calculate the initial amounts of H _{2}PO_{3}^{-} and NaOH in moles before the reaction happens.**

Phosphorous acid, H_{3}PO_{4}(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.

pK_{a1} = 1.30

pK_{a2} = 6.70

Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.5 M H_{3}PO_{3}(aq) with 1.5 M KOH(aq).

Part E. after addition of 100.0 mL of KOH

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