Chapter 10, Problem 10.24P

(a)

Interpretation Introduction

Interpretation:

Which $\text{pH}$ both $\left[{\text{H}}_{\text{2}}\text{A}\right]\text{=}\left[{\text{HA}}^{\text{-}}\right]$ are equal has to be identified.

Concept introduction:

Henderson-Hassel Balch equation

$\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{1}}\text{+log}\frac{\left[{\text{HA}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}\text{A}\right]}\\ \text{pH}\text{=}{\text{pK}}_{\text{2}}\text{+log}\frac{\left[{\text{A}}^{\text{2-}}\right]}{\left[{\text{HA}}^{\text{-}}\right]}\end{array}$

The $\text{pH}$ of solution can be calculated as,

$\text{pH=}\text{-log}{\left[{\text{H}}^{\text{+}}\right]}_{\text{γ}}{}_{{\text{H}}^{\text{-}}}$

With the above equation, the negative logarithm of activity of Hydrogen ion can be measured.

(b)

Interpretation Introduction

Interpretation:

Which $\text{pH}$ both $\left[{\text{HA}}^{\text{-}}\right]\text{=}\left[{\text{A}}^{\text{2-}}\right]$ are equal has to be identified.

Concept introduction:

Henderson-Hassel Balch equation

$\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{1}}\text{+log}\frac{\left[{\text{HA}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}\text{A}\right]}\\ \text{pH}\text{=}{\text{pK}}_{\text{2}}\text{+log}\frac{\left[{\text{A}}^{\text{2-}}\right]}{\left[{\text{HA}}^{\text{-}}\right]}\end{array}$

The $\text{pH}$ of solution can be calculated as,

$\text{pH=}\text{-log}{\left[{\text{H}}^{\text{+}}\right]}_{\text{γ}}{}_{{\text{H}}^{\text{-}}}$

With the above equation, the negative logarithm of activity of Hydrogen ion can be measured.

(c)

Interpretation Introduction

Interpretation:

The principal species at $\text{pH}\text{=}\text{2}\text{.00:}{\text{H}}_{\text{2}}\text{A}{\text{,HA}}^{\text{-}}\text{,or}{\text{A}}^{\text{2-}}$ has to be explained.

(d)

Interpretation Introduction

Interpretation:

The principal species at $\text{pH}\text{=}6.\text{00}$ has to be explained.

(e)

Interpretation Introduction

Interpretation:

Which the principal species is at $\text{pH}\text{=}6.\text{00}$ has to be explained.