Chapter 16, Problem 16.17QP

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given solutions has to be calculated

Concept Information:

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

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 equation of equilibrium for autoionization of water is,

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

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

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.

The $\text{pH}$ of the given solutions

Answer to Problem 16.17QP

The $\text{pH}$ of the given solution (a) is 3.00

Explanation of Solution

Calculation of hydrogen ion concentration:

$\text{HCl}$ is a strong acid, so the concentration of hydrogen ion is also $0.0010M$

Calculation of $\text{pH}$

The $\text{pH}$ is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(0}\text{.0010})\\ =3.00\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given solutions has to be calculated

Concept Information:

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

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 equation of equilibrium for autoionization of water is,

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

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

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.

Answer to Problem 16.17QP

The $\text{pH}$ of the given solution (b) is 13.89

Explanation of Solution

Calculation of hydrogen ion concentration:

$\text{KOH}$ is an ionic compound and gets fully ionized in water.

The hydrogen ion concentration can be found as follows,

$\begin{array}{l}\text{[pOH]}\text{=}-\log {\text{[OH}}^{\text{-}}\text{]}\\ =-\log \text{[0}\text{.76]}\\ \text{[pOH]}=0.119\end{array}$

Calculation of $\text{pH}$

The $\text{pH}$ is calculated as follows,

$\begin{array}{l}\text{pH +}\text{pOH = 14}\text{.00}\\ \text{pH =14}\text{.00 - pOH}\\ \text{=}\text{14}\text{.00 -}\text{0}\text{.119 }\\ \text{pH}\text{=}\text{13}\text{.88}\\ =13.89\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given solutions has to be calculated

Concept Information:

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

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 equation of equilibrium for autoionization of water is,

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

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

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.

The $\text{pH}$ of the given solutions

Answer to Problem 16.17QP

The $\text{pH}$ of the given solution (c) is $10.75$

Explanation of Solution

Calculation of hydrogen ion concentration:

${\text{Ba(OH)}}_{\text{2}}$ is ionic and gets fully ionized in water.  The concentration of the hydroxide ion is $2.8\times {10}^{-4}$

The hydrogen ion concentration can be found as follows,

$\begin{array}{l}\text{[pOH]}\text{=}-\log {\text{[OH}}^{\text{-}}\text{]}\\ =-\log \text{[2}\times \text{2}\text{.8}\times {\text{10}}^{\text{-4}}\text{]}\\ =-\log \text{[}5.6\times {\text{10}}^{\text{-4}}\text{]}\\ \text{=}\text{3}\text{.251}\\ \text{[pOH]}=\text{3}\text{.251}\end{array}$

Calculation of $\text{pH}$

The $\text{pH}$ is calculated as follows,

$\begin{array}{l}\text{pH +}\text{pOH = 14}\text{.00}\\ \text{pH =14}\text{.00 - pOH}\\ \text{=}\text{14}\text{.00 -}3.25\\ \text{pH}\text{=}10.75\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given solutions has to be calculated

Concept Information:

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

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 equation of equilibrium for autoionization of water is,

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

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

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.

The $\text{pH}$ of the given solutions

Answer to Problem 16.17QP

The $\text{pH}$ of the given solution (d) is 3.28

Explanation of Solution

Calculation of hydrogen ion concentration:

Nitric acid is a strong acid, so the concentration of hydrogen ion is also $5.2\times {10}^{-4}M$

Calculation of $\text{pH}$

The $\text{pH}$ is calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(5}\text{.2}\times {\text{10}}^{\text{-4}})\\ =3.28\end{array}$