Chapter 9, Problem 9.BE

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}\text{HBr}$ (without activites) has to be calculated.

Concept Introduction:

Mass balance: Mass balance can be defined as the amount of all species that is present in a solution having a specific atom that should be same as the quantity of that atom that is given to the solution.

Charge balance: Charge balance can be defined as the sum of all positive ions present in solution will be same as the sum of all negative ions present in the solution.

pH calculation:

The relationship between $\text{pH}\text{and}\text{pOH}$ is given as,

${\text{pH+pOH=-log K}}_{\text{w}}$

$\text{pH+pOH=14}$

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

But,

$\begin{array}{l}{\text{A}}_{{\text{H}}^{\text{+}}}{\text{+A}}_{{\text{OH}}^{\text{-}}}{\text{=K}}_{\text{w}}\\ \\ {\text{A}}_{{\text{H}}^{\text{+}}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{A}}_{{\text{OH}}^{\text{-}}}}\end{array}$

$\begin{array}{l}{\text{pH=-logA}}_{{\text{H}}^{\text{+}}}\\ \\ \text{pH=-log}{\left[{\text{H}}^{\text{+}}\right]}_{{\text{γH}}^{\text{+}}}\end{array}$

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

Answer to Problem 9.BE

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}\text{HBr}$ is found to be $\text{6}\text{.98}$.

Explanation of Solution

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

Mass balance of $\text{HBr}$= $\left[{\text{Br}}^{\text{-}}\right]\text{=1}{\text{.0×10}}^{\text{-8}}\text{M}$

Equilibrium = $\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{w}}$

Take ,

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=x}\\ \\ \left[{\text{Br}}^{\text{-}}\right]\text{=1}{\text{.0×10}}^{\text{-8}}\text{M}\end{array}$

Based on charge balance , $\left[{\text{OH}}^{\text{-}}\right]\text{=x-1}{\text{.0×10}}^{\text{-8}}\text{M}$

$\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{w}}$

Substitute $\left[{\text{OH}}^{\text{-}}\right]\text{=x-1}{\text{.0×10}}^{\text{-8}}\text{M}$ in ${\text{K}}_{\text{w}}$

$\left(\text{x}\right)\left(\text{x-1}{\text{.0×10}}^{\text{-8}}\right)\text{=1}{\text{.0×10}}^{\text{-14}}$

$\text{x=1}{\text{.0}}_{\text{5}}{\text{×10}}^{\text{-7}}\text{M}$

$\text{pH}$ can be calculated as,

$\begin{array}{l}\text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\\ \\ \text{pH=-log}\left(\text{1}{\text{.0}}_{\text{5}}{\text{×10}}^{\text{-7}}\text{M}\right)\\ \\ \text{pH=6}\text{.98}\end{array}$

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}\text{HBr}$ = $\text{6}\text{.98}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ (without activites) has to be calculated.

Concept Introduction:

Mass balance: Mass balance can be defined as the amount of all species that is present in a solution having a specific atom that should be same as the quantity of that atom that is given to the solution.

Charge balance: Charge balance can be defined as the sum of all positive ions present in solution will be same as the sum of all negative ions present in the solution.

pH calculation:

The relationship between $\text{pH}\text{and}\text{pOH}$ is given as,

${\text{pH+pOH=-log K}}_{\text{w}}$

$\text{pH+pOH=14}$

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

But,

$\begin{array}{l}{\text{A}}_{{\text{H}}^{\text{+}}}{\text{+A}}_{{\text{OH}}^{\text{-}}}{\text{=K}}_{\text{w}}\\ \\ {\text{A}}_{{\text{H}}^{\text{+}}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{A}}_{{\text{OH}}^{\text{-}}}}\end{array}$

$\begin{array}{l}{\text{pH=-logA}}_{{\text{H}}^{\text{+}}}\\ \\ \text{pH=-log}{\left[{\text{H}}^{\text{+}}\right]}_{{\text{γH}}^{\text{+}}}\end{array}$

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

Answer to Problem 9.BE

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is found to be $\text{6}\text{.96}$ .

Explanation of Solution

Charge balance of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ = $\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+2}\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$

Mass balance of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$= $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]\text{=1}{\text{.0×10}}^{\text{-8}}\text{M}$

Equilibrium = $\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{w}}$

Take ,

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=x}\\ \\ \left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]\text{=1}{\text{.0×10}}^{\text{-8}}\text{M}\end{array}$

Based on charge balance , $\left[{\text{OH}}^{\text{-}}\right]\text{=x-1}{\text{.0×10}}^{\text{-8}}\text{M}$

$\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{w}}$

Substitute $\left[{\text{OH}}^{\text{-}}\right]\text{=x-2}{\text{.0×10}}^{\text{-8}}\text{M}$ in ${\text{K}}_{\text{w}}$

$\left(\text{x}\right)\left(\text{x-2}{\text{.0×10}}^{\text{-8}}\right)\text{=1}{\text{.0×10}}^{\text{-14}}$

$\text{x=1}{\text{.10×10}}^{\text{-7}}\text{M}$

$\text{pH}$ can be calculated as,

$\begin{array}{l}\text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\\ \\ \text{pH=-log}\left(\text{1}{\text{.10×10}}^{\text{-7}}\text{M}\right)\\ \\ \text{pH=6}\text{.96}\end{array}$

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ = $\text{6}\text{.96}$