Chapter 11, Problem 11.14P

(a)

Interpretation Introduction

Interpretation:

Concentration of each species in below diagram has to be calculated.

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged for of equilibrium constant, $K_{\text{a}}$. Consider an equation of dissociation of an acid as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{pH}$ is as follows:

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

Here,

$p{K}_{\text{a}}$ is acid dissociation constant of weak acid $\text{HA}$.

$\left[{\text{A}}^{-}\right]$ is concentration of conjugate base ${\text{A}}^{-}$

$\left[\text{HA}\right]$ is concentration of acid $\text{HA}$.

Explanation of Solution

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_1+\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$        (1)

Substitute $\text{p}{K}_1$ for $\text{pH}$ in equation (1).

  $\begin{array}{c}\text{p}{K}_1=\text{p}{K}_1+\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}\\ \log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}=0\\ \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}=1\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_1$, concentration of ${\text{HA}}^{-}$ is equal to ${\text{H}}_2\text{A}$.

In accordance to Henderson Hasselbach equation ${\text{pH}}_1$ is calculated as follows:

  ${\text{pH}}_1=\text{p}{K}_1+\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$        (1)

In accordance to Henderson Hasselbach equation ${\text{pH}}_2$ is calculated as follows:

  ${\text{pH}}_2=\text{p}{K}_2+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}$        (2)

The expression for final $\text{pH}$ is calculated as follows:

  $\text{pH}=\frac{1}{2}\left({\text{pH}}_1+{\text{pH}}_2\right)$        (3)

Substitute $\text{p}{K}_1+\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$ for ${\text{pH}}_1$ and $\text{p}{K}_2+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}$ for ${\text{pH}}_2$ in equation (3).

  $\begin{array}{c}\text{pH}=\frac{1}{2}\left(\text{p}{K}_1+\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}+\text{p}{K}_2+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)+\frac{1}{2}\left(\log \frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)+\frac{1}{2}\log \left(\frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{H}}_2\text{A}\right]}\right)\end{array}$

Substitute $\text{pH}$ for $\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)$ in above equation.

  $\begin{array}{c}\text{pH}=\text{pH}+\frac{1}{2}\log \left(\frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{H}}_2\text{A}\right]}\right)\\ \log \left(\frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{H}}_2\text{A}\right]}\right)=0\\ \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{H}}_2\text{A}\right]}=1\end{array}$

At $\text{pH}$ equal to $\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)$, concentration of ${\text{A}}^{2-}$ is equal to ${\text{H}}_2\text{A}$.

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_2+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}$        (2)

Substitute $\text{p}{K}_2$ for $\text{pH}$ in equation (2).

  $\begin{array}{c}\text{p}{K}_2=\text{p}{K}_2+\log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}\\ \log \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}=0\\ \frac{\left[{\text{A}}^{2-}\right]}{\left[{\text{HA}}^{-}\right]}=1\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_2$, concentration of ${\text{A}}^{2-}$ is equal to ${\text{HA}}^{-}$.

(b)

Interpretation Introduction

Interpretation:

Analogous diagrams for monoprotic and triprotic systems have to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

In diagram of monoprotic system if $\text{pH}$ is equal to $\text{p}{K}_{\text{a}}$ then solution is considered as neutral. If value of $\text{pH}$ is less than $\text{p}{K}_{\text{a}}$, solution becomes acidic and value of $\text{pH}$ is more than $\text{p}{K}_{\text{a}}$, solution becomes basic. Hence, diagram is as follows:

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_1+\log \frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$        (4)

Substitute $\text{p}{K}_1$ for $\text{pH}$ in equation (4).

  $\begin{array}{c}\text{p}{K}_1=\text{p}{K}_1+\log \frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\\ \log \frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=0\\ \frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=1\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_1$, concentration of ${\text{A}}^{-}$ is equal to $\text{HA}$.

In diagram of triprotic system diagram consist of $\text{p}{K}_1$, $\text{p}{K}_2$ and $\text{p}{K}_3$ that is related to $\text{pH}$ in acidic, neutral and basic region. Hence, diagram is as follows:

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_1+\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$        (5)

Substitute $\text{p}{K}_1$ for $\text{pH}$ in equation (5).

  $\begin{array}{c}\text{p}{K}_1=\text{p}{K}_1+\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}\\ \log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}=0\\ \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}=1\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_1$, concentration of ${\text{H}}_2{\text{A}}^{-}$ is equal to ${\text{H}}_3\text{A}$.

In accordance to Henderson Hasselbach equation ${\text{pH}}_1$ is calculated as follows:

  ${\text{pH}}_1=\text{p}{K}_1+\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$        (6)

In accordance to Henderson Hasselbach equation ${\text{pH}}_2$ is calculated as follows:

  ${\text{pH}}_2=\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$        (7)

The expression for final $\text{pH}$ is calculated as follows:

  $\text{pH}=\frac{1}{2}\left({\text{pH}}_1+{\text{pH}}_2\right)$        (8)

Substitute $\text{p}{K}_1+\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}$ for ${\text{pH}}_1$ and $\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$ for ${\text{pH}}_2$ in equation (8).

  $\begin{array}{c}\text{pH}=\frac{1}{2}\left(\text{p}{K}_1+\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}+\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)+\frac{1}{2}\left(\log \frac{\left[{\text{H}}_2{\text{A}}^{-}\right]}{\left[{\text{H}}_3\text{A}\right]}+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)+\frac{1}{2}\log \left(\frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}\right)\end{array}$

Substitute $\text{pH}$ for $\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)$ in above equation.

  $\begin{array}{c}\text{pH}=\text{pH}+\frac{1}{2}\log \left(\frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}\right)\\ \log \left(\frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}\right)=0\\ \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}=1\\ \left[{\text{HA}}^{2-}\right]=\left[{\text{H}}_3\text{A}\right]\end{array}$

At $\text{pH}$ equal to $\frac{1}{2}\left(\text{p}{K}_1+\text{p}{K}_2\right)$, concentration of ${\text{HA}}^{2-}$ is equal to ${\text{H}}_3\text{A}$.

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$        (9)

Substitute $\text{p}{K}_2$ for $\text{pH}$ in equation (9).

  $\begin{array}{c}\text{p}{K}_2=\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\\ \log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}=0\\ \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}=1\\ \left[{\text{HA}}^{2-}\right]=\left[{\text{H}}_2{\text{A}}^{-}\right]\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_2$, concentration of ${\text{HA}}^{2-}$ is equal to ${\text{H}}_2{\text{A}}^{-}$.

In accordance to Henderson Hasselbach equation ${\text{pH}}_2$ is calculated as follows:

  ${\text{pH}}_2=\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}$        (10)

In accordance to Henderson Hasselbach equation ${\text{pH}}_3$ is calculated as follows:

  ${\text{pH}}_3=\text{p}{K}_3+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}$        (11)

The expression for final $\text{pH}$ is calculated as follows:

  $\text{pH}=\frac{1}{2}\left({\text{pH}}_2+{\text{pH}}_3\right)$        (12)

Substitute $\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}$ for ${\text{pH}}_2$ and $\text{p}{K}_3+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}$ for ${\text{pH}}_3$ in equation (12).

  $\begin{array}{c}\text{pH}=\frac{1}{2}\left(\text{p}{K}_2+\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_3\text{A}\right]}+\text{p}{K}_3+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_2+\text{p}{K}_3\right)+\frac{1}{2}\left(\log \frac{\left[{\text{HA}}^{2-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}\right)\\ =\frac{1}{2}\left(\text{p}{K}_2+\text{p}{K}_3\right)+\frac{1}{2}\log \left(\frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\right)\end{array}$

Substitute $\text{pH}$ for $\frac{1}{2}\left(\text{p}{K}_2+\text{p}{K}_3\right)$ in above equation.

  $\begin{array}{c}=\text{pH}+\frac{1}{2}\log \left(\frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\right)\\ \log \left(\frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}\right)=0\\ \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{H}}_2{\text{A}}^{-}\right]}=1\\ \left[{\text{A}}^{3-}\right]=\left[{\text{H}}_2{\text{A}}^{-}\right]\end{array}$

At $\text{pH}$ equal to $\frac{1}{2}\left(\text{p}{K}_2+\text{p}{K}_3\right)$, concentration of ${\text{A}}^{3-}$ is equal to ${\text{H}}_2{\text{A}}^{-}$.

In accordance to Henderson Hasselbach equation $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_3+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}$        (13)

Substitute $\text{p}{K}_3$ for $\text{pH}$ in equation (13).

  $\begin{array}{c}\text{p}{K}_3=\text{p}{K}_3+\log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}\\ \log \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}=0\\ \frac{\left[{\text{A}}^{3-}\right]}{\left[{\text{HA}}^{2-}\right]}=1\\ \left[{\text{A}}^{3-}\right]=\left[{\text{HA}}^{2-}\right]\end{array}$

At $\text{pH}$ equal to $\text{p}{K}_3$, concentration of ${\text{A}}^{3-}$ is equal to ${\text{HA}}^{2-}$.