Chapter 16, Problem 16.38QP

Calculate the pOH and pH of the following aqueous solutions at 25°C: (a) 0.016 M KOH, (b) 1.35 M NaOH, (c) 0.094 M Ba(OH)₂.

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete.  This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

Strong bases:

Strong bases are formed from alkali metals and alkaline earth metals of Group IA and IIA respectively.

Strong base dissociates into its constituent ions.

For Group IA metal hydroxides, the hydroxide ion concentration is simply the initial concentration of the strong base

For Group IIA metal hydroxides, the hydroxide ion concentration at equilibrium will be twice that of the initial concentration of strong base

$\text{pOH}$ definition

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$

The $\text{pH}$ and $\text{pOH}$ of the given $0.016M\text{KOH}$ solution

Answer to Problem 16.38QP

Answer

The $\text{pH}$ and $\text{pOH}$ of the given solution (a) is 12.20 and 1.80  respectively

Explanation of Solution

Concentration of  ${\text{[OH}}^{\text{-}}\text{]}$

In a strong base solution, the molar concentration of hydroxide ion is equal to the base concentration.

$\text{KOH}$ is a strong base (monobasic).  Therefore, the hydroxide ion concentration will be equal to that of  $\text{KOH}$ concentration.

${\text{[OH}}^{\text{-}}\text{]}\text{=}\text{[KOH}]=0.016M$

Calculation of $\text{pOH}$

The $\text{pOH}$ of the given solution can be calculated from the concentration of hydroxide ion as follows,

$\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(0}\text{.016)}\\ \text{=}\text{1}\text{.80}\end{array}$

Calculation of $\text{pH}$

The relationship between $\text{pH}$ and $\text{pOH}$ can be given by the following equation

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

$\begin{array}{l}\text{On rearranging,}\\ \text{pH}=14.00-\text{pOH}\\ =14.00-1.80\\ =12.20\end{array}$

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete.  This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

Strong bases:

Strong bases are formed from alkali metals and alkaline earth metals of Group IA and IIA respectively.

Strong base dissociates into its constituent ions.

For Group IA metal hydroxides, the hydroxide ion concentration is simply the initial concentration of the strong base

For Group IIA metal hydroxides, the hydroxide ion concentration at equilibrium will be twice that of the initial concentration of strong base

$\text{pOH}$ definition

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$

The $\text{pH}$ and $\text{pOH}$ of the given $1.35M\text{NaOH}$ solution

Answer to Problem 16.38QP

Answer

The $\text{pH}$ and $\text{pOH}$ of the given solution (b) is 14.13 and 0.13  respectively

Explanation of Solution

Concentration of ${\text{[OH}}^{\text{-}}\text{]}$

In a strong base solution, the molar concentration of hydroxide ion is equal to the base concentration.

$\text{NaOH}$ is a strong base (monobasic).  Therefore, the hydroxide ion concentration will be equal to that of $\text{NaOH}$ concentration.

${\text{[OH}}^{\text{-}}\text{]}\text{=}\text{[NaOH}]=1.35M$

Calculation of $\text{pOH}$

The $\text{pOH}$ of the given solution can be calculated from the concentration of hydroxide ion as follows,

$\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(1}\text{.35)}\\ \text{=}-0.13\end{array}$

Calculation of $\text{pH}$

The relationship between $\text{pH}$ and $\text{pOH}$ can be given by the following equation

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

$\begin{array}{l}\text{On rearranging,}\\ \text{pH}=14.00-\text{pOH}\\ =14.00-(-0.13)\\ =14.13\end{array}$

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete.  This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

Strong bases:

Strong bases are formed from alkali metals and alkaline earth metals of Group IA and IIA respectively.

Strong base dissociates into its constituent ions.

For Group IA metal hydroxides, the hydroxide ion concentration is simply the initial concentration of the strong base

For Group IIA metal hydroxides, the hydroxide ion concentration at equilibrium will be twice that of the initial concentration of strong base

$\text{pOH}$ definition

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The $\text{pH}$ and $\text{pOH}$ of the given aqueous solutions at ${25}^{\circ }\text{C}$

The $\text{pH}$ and $\text{pOH}$ of the given $0.094M\text{Ba}{(\text{OH)}}_{\text{2}}$ solution

Answer to Problem 16.38QP

Answer

The $\text{pH}$ and $\text{pOH}$ of the given solution (c) is 13.27 and 0.73  respectively

Explanation of Solution

Concentration of  ${\text{[OH}}^{\text{-}}\text{]}$

$\text{Ba}{(\text{OH)}}_{\text{2}}$ is a strong base (dibasic).  Barium belongs to the Group IIA.  Therefore, the hydroxide ion concentration will be twice that of the $\text{Ba}{(\text{OH)}}_{\text{2}}$ concentration.

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}2\times {\text{[Ba(OH)}}_{\text{2}}]\\ =2\times 0.094M\\ =0.188M\end{array}$

Calculation of $\text{pOH}$

The $\text{pOH}$ of the given solution can be calculated from the concentration of hydroxide ion as follows,

$\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(0}\text{.188)}\\ \text{=}0.73\end{array}$

Calculation of $\text{pH}$

The relationship between $\text{pH}$ and $\text{pOH}$ can be given by the following equation

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

$\begin{array}{l}\text{On rearranging,}\\ \text{pH}=14.00-\text{pOH}\\ =14.00-0.73\\ =13.27\end{array}$