Chapter 11, Problem 11.72P

Interpretation Introduction

Interpretation:

The graph for the titration of triprotic with strong base, $\text{50}\text{.0}\text{mL}$ of $\text{0}\text{.0200}\text{M}\text{histidine}$ with $\text{2HCl}$ of $\text{0}\text{.100}\text{M}\text{NaOH}$ was plotted.

Concept Introduction:-

Titration of triprotic acid with strong base:

When a strong acid is titrated against triprotic acid, there is an equivalence or stoichiometric point, when an H⁺ is dissociated then there are multiple equivalence points. The equivalence point will occur when the moles of the strong base added to same amount of the moles of weak acid present. However, when a weak polyprotic acid is titrated. Therefore, the number of equivalence points depends on the number of H⁺ atoms that can be removed from the molecule.

Spreadsheet Equation for titration of triprotic with strong base.

$\varphi \equiv \frac{{\text{C}}_{\text{b}}{\text{V}}_{\text{b}}}{{\text{C}}_{\text{a}}{\text{V}}_{\text{a}}}\text{=}\frac{{\text{α}}_{{\text{H}}_{\text{2}}{\text{A}}^{\text{-}}}{\text{+2α}}_{{\text{HA}}^{\text{2-}}}{\text{+3α}}_{{\text{A}}^{\text{3}}}\text{-}\frac{\left[{\text{H}}^{\text{+}}\right]\text{-}\left[{\text{OH}}^{\text{-}}\right]}{{\text{C}}_{\text{a}}}}{\text{1+}\frac{\left[{\text{H}}^{\text{+}}\right]\text{-}\left[{\text{OH}}^{\text{-}}\right]}{{\text{C}}_{\text{b}}}}$

Where,

${\text{C}}_{\text{a}}$=initial concentration

${\text{C}}_b$=final concentration

${\text{V}}_{\text{a}}$=volume of the acid

${\text{V}}_{\text{b}}$= volume of the base

${\text{α}}_{\text{A}}$ =acid dissociation constant.

$\varphi$ =fraction way to equivalence point.