Chapter 15, Problem 15.114QP

The ion product of D₂O is 1.35 × 10⁻¹⁵ at 25°C. (a) Calculate pD where pD = −log [D⁺]. (b) For what values of pD will a solution be acidic in D₂O? (c) Derive a relation between pD and pOD.

Interpretation Introduction

Interpretation:

The pD for ${\text{D}}_{\text{2}}\text{O}$ has to be calculated

For what values of pD will a solution be acidic in ${\text{D}}_{\text{2}}\text{O}$ has to be explained

A relation between pD and pOD has to be derived.

Concept Information:

Autoionization of water:

The equation of equilibrium for autoionization of water is,

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$

Relationship between pH and 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.

$\text{pH}$ 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.

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

To calculate: The pD for ${\text{D}}_{\text{2}}\text{O}$

Answer to Problem 15.114QP

The pD value is 7.43

Explanation of Solution

Record the given datas

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is $1.35\times {10}^{-15}$ at ${25}^{\circ }\text{C}$

$\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is given as $1.35\times {10}^{-15}$. The formula for calculation pD is given as $\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

Autoionization of ${\text{D}}_{\text{2}}\text{O}$

The autoionization of deuterium substituted water is ${\text{D}}_{\text{2}}\text{O}\rightleftarrows {\text{D}}^{\text{+}}\text{+}{\text{OD}}^{\text{-}}$

${\text{[D}}^{\text{+}}{\text{][OD}}^{\text{-}}\text{]=1}\text{.35}\times {\text{10}}^{\text{-15}}$

From the given data,

The definition of pD is $\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

pD can be calculated as follows,

$\begin{array}{l}\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}\\ \text{=-log}\sqrt{1.35\times {10}^{-15}}\\ =7.43\end{array}$

Therefore, the pD is 7.43

Interpretation Introduction

Interpretation:

The pD for ${\text{D}}_{\text{2}}\text{O}$ has to be calculated

For what values of pD will a solution be acidic in ${\text{D}}_{\text{2}}\text{O}$ has to be explained

A relation between pD and pOD has to be derived.

Concept Information:

Autoionization of water:

The equation of equilibrium for autoionization of water is,

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$

Relationship between pH and 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.

$\text{pH}$ 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.

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

Answer to Problem 15.114QP

The solution to be acidic, pD must be $<7.43$

Explanation of Solution

Record the given datas

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is $1.35\times {10}^{-15}$ at ${25}^{\circ }\text{C}$

$\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is given as $1.35\times {10}^{-15}$. The formula for calculation pD is given as $\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

Autoionization of ${\text{D}}_{\text{2}}\text{O}$

The autoionization of deuterium substituted water is ${\text{D}}_{\text{2}}\text{O}\rightleftarrows {\text{D}}^{\text{+}}\text{+}{\text{OD}}^{\text{-}}$

${\text{[D}}^{\text{+}}{\text{][OD}}^{\text{-}}\text{]=1}\text{.35}\times {\text{10}}^{\text{-15}}$

To be acidic, the pD must be $<7.43$

Interpretation Introduction

Interpretation:

The pD for ${\text{D}}_{\text{2}}\text{O}$ has to be calculated

For what values of pD will a solution be acidic in ${\text{D}}_{\text{2}}\text{O}$ has to be explained

A relation between pD and pOD has to be derived.

Concept Information:

Autoionization of water:

The equation of equilibrium for autoionization of water is,

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$

Relationship between pH and 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.

$\text{pH}$ 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.

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

To Derive: A relation between pD and pOD

Answer to Problem 15.114QP

The relation between pD and pOD is 14.87

Explanation of Solution

Record the given datas

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is $1.35\times {10}^{-15}$ at ${25}^{\circ }\text{C}$

$\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

The ion product of ${\text{D}}_{\text{2}}\text{O}$ is given as $1.35\times {10}^{-15}$. The formula for calculation pD is given as $\text{pD}\text{=}{\text{-log[D}}^{\text{+}}\text{]}$

Autoionization of ${\text{D}}_{\text{2}}\text{O}$

The autoionization of deuterium substituted water is ${\text{D}}_{\text{2}}\text{O}\rightleftarrows {\text{D}}^{\text{+}}\text{+}{\text{OD}}^{\text{-}}$

${\text{[D}}^{\text{+}}{\text{][OD}}^{\text{-}}\text{]=1}\text{.35}\times {\text{10}}^{\text{-15}}$

From autoionization of ${\text{D}}_{\text{2}}\text{O}$ we know that,

$\begin{array}{l}{\text{[D}}^{\text{+}}{\text{][OD}}^{\text{-}}\text{]=1}\text{.35}\times {\text{10}}^{\text{-15}}\\ \text{Taking -log on both sides of the equation, we get}\\ {\text{-log[D}}^{\text{+}}{\text{]+-log[OD}}^{\text{-}}\text{]=-log(1}\text{.35}\times {\text{10}}^{\text{-15}})\\ \text{pD+pOD=14}\text{.87}\end{array}$

Therefore, the relation between pD and pOD is $\text{pD+pOD=14}\text{.87}$