Chapter 11, Problem 11.1P

Interpretation Introduction

Interpretation:

Chemistry that governs $\text{pH}$ at each indicated point on given curve has to be stated.

Concept Introduction:

$\text{pH}$ is calculated by Henderson-HasselBalch equation given as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Here,

$\text{p}{K}_{\text{a}}$ denotes characteristic acid dissociation constant of diprotic acid.

Explanation of Solution

Initial point at A corresponds to weak acid ${\text{H}}_2\text{A}$. Thus formal concentration of weak acid ${\text{H}}_2\text{A}$ is used at this point to compute $\text{pH}$.

Point B corresponds to first buffer-pair of ${\text{HA}}^{-}$ and ${\text{H}}_2\text{A}$. Thus Henderson-Hasselbalch equation governs the chemistry at this point.

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}\right)$

Point C represents first equivalence point corresponding to loss of proton from ${\text{H}}_2\text{A}$ to form ${\text{HA}}^{-}$. Since ${\text{HA}}^{-}$ formed at this stage is an intermediate form of ${\text{H}}_2\text{A}$, so its $\text{pH}$ is governed by equation as follows:

  $\text{pH}\approx \frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)$

Point D corresponds to second buffer-pair of ${\text{HA}}^{-}$ and ${\text{A}}^{2-}$. Thus Henderson-Hasselbalch equation that governs the chemistry at this point is as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}\right)$

Point E represents second equivalence point corresponding to loss of proton from ${\text{HA}}^{-}$ to form ${\text{A}}^{2-}$. Since all ${\text{HA}}^{-}$ has converted to ${\text{A}}^{2-}$, $\text{pH}$ formed at this stage is determined from treatment of ${\text{A}}^{2-}$ as monoprotic base.

Beyond E there is excess of ${\text{OH}}^{-}$ due to reaction of ${\text{A}}^{2-}$ with water to form ${\text{OH}}^{-}$. Excess ${\text{OH}}^{-}$ gives $\text{pOH}$ that further gives $\text{pH}$.