Chapter 9, Problem 9.45P

a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution has to be calculated and compared.

Concept Introduction:

The concentrations of $\text{HA}$ and ${\text{A}}^{\text{-}}$ will not be equal to their formal concentrations for dilute solutions or at extreme $\text{pH}$ .

Consider the mixing of ${\text{F}}_{\text{HA}}$ moles of $\text{HA}$ and ${\text{F}}_{{\text{A}}^{\text{-}}}$ moles of the salt ${\text{Na}}^{\text{+}}{\text{A}}^{\text{-}}$. Then, the mass and charge balances can be given as,

Mass balance: ${\text{F}}_{\text{HA}}{\text{+F}}_{{\text{A}}^{\text{-}}}\text{=}\left[\text{HA}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]$

Charge balance: $\left[{\text{Na}}^{\text{+}}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]$

The substitution $\left[{\text{Na}}^{\text{+}}\right]{\text{=F}}_{{\text{A}}^{\text{-}}}$ leads to the equation,

$\begin{array}{l}\left[\text{HA}\right]{\text{=F}}_{\text{HA}}\text{-}\left[{\text{H}}^{\text{+}}\right]\text{+}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{A}}^{\text{-}}\right]{\text{=F}}_{{\text{A}}^{\text{-}}}\text{+}\left[{\text{H}}^{\text{+}}\right]\text{-}\left[{\text{OH}}^{\text{-}}\right]\end{array}$

Assumption has to be made that $\left[\text{HA}\right]\approx {\text{F}}_{\text{HA}}$ and $\left[{\text{A}}^{\text{-}}\right]\approx {\text{F}}_{{\text{A}}^{\text{-}}}$ and these values are used in Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where , ${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

The value of ${\text{K}}_{\text{w}}$ is calculated by the formula,

${\text{K}}_{\text{w}}{\text{=K}}_{\text{a}}{\text{×K}}_{\text{b}}$

The $\text{pH}$ of weak base is calculated using the equation,

$\begin{array}{l}\text{pH=}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\end{array}$

Answer to Problem 9.45P

The $\text{pH}$ of the solution is $\text{11}\text{.48}$ .

Explanation of Solution

Mass balance and Charge balance is obtained if we dissolve $\text{B}$ and ${\text{BH}}^{\text{+}}{\text{Br}}^{\text{-}}$ .

Mass balance = ${\text{F}}_{\text{BH}}{\text{+F}}_{\text{B}}\text{=}\left[{\text{BH}}^{\text{+}}\right]\text{+}\left[\text{B}\right]$

Charge balance = $\left[{\text{Br}}^{\text{-}}\right]\text{+}\left[{\text{OH}}^{\text{-}}\right]\text{=}\left[{\text{BH}}^{\text{+}}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]$

If, $\left[{\text{Br}}^{\text{-}}\right]{\text{=F}}_{{\text{BH}}^{\text{+}}}$ , then the charge balance is written as,

$\left[{\text{BH}}^{\text{+}}\right]{\text{=F}}_{{\text{BH}}^{\text{+}}}\text{+}\left[{\text{OH}}^{\text{-}}\right]\text{-}\left[{\text{H}}^{\text{-}}\right]$ (1)

Substituting the above expression into the mass balance gives,

$\left[\text{B}\right]{\text{=F}}_{\text{B}}\text{-}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]$ (2)

Assume that

$\begin{array}{l}\left[\text{B}\right]\text{=0}\text{.0100}\text{M}\\ \\ \left[{\text{BH}}^{\text{+}}\right]\text{=0}\text{.0200}\text{M}\end{array}$

Then, the $\text{pH}$ is calculated as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

$\begin{array}{l}\text{pH=12}\text{.00+log}\frac{\text{0}\text{.0100}}{\text{0}\text{.0200}}\\ \\ \text{pH=11}\text{.70}\end{array}$

If the given below values are not assumed, then equations (1) and (2) can be used

$\begin{array}{l}\left[\text{B}\right]\text{=0}\text{.0100}\text{M}\\ \\ \left[{\text{BH}}^{\text{+}}\right]\text{=0}\text{.0200}\text{M}\end{array}$

The solution is basic and $\left[{\text{H}}^{\text{+}}\right]$ are neglected relative to $\left[{\text{OH}}^{\text{-}}\right]$ and then $\left[\text{B}\right]$ is written as,

$\begin{array}{l}\left[\text{B}\right]\text{=0}\text{.0100-x}\\ \\ \left[{\text{BH}}^{\text{+}}\right]\text{=0}\text{.0200+x}\end{array}$

Where x= $\left[{\text{OH}}^{\text{-}}\right]$

${\text{K}}_{\text{b}}$ takes the value of ${\text{10}}^{\text{-2}\text{.00}}$ = $\frac{\left[{\text{BH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[\text{B}\right]}\text{=}\frac{\left(\text{0}\text{.0200+x}\right)\left(\text{x}\right)}{\left(\text{0}\text{.0100-x}\right)}$

$\text{x=0}\text{.00303}\text{M}$

$\text{pH=-log}\frac{{\text{K}}_{\text{w}}}{\text{x}}$

$\text{pH=11}\text{.48}$

The $\text{pH}$ of the solution = $\text{11}\text{.48}$

b)

Interpretation Introduction

Interpretation:

The goal seek has to be done for the calculations done in part A.

Answer to Problem 9.45P

Figure 1

Explanation of Solution

The goal seek is used to vary cell $\text{B5}$ till cell $\text{D4}$ that is equal to ${\text{K}}_{\text{a}}$.

Figure 1