Chapter 9, Problem 9.81P

Interpretation Introduction

(a)

Interpretation:

Themissing value of ${\text{[OH}}^{-}\text{]}$ and $\text{pH}$ for the given solution with $\text{5}\text{.3 }\times {\text{10}}^{\text{-3}}$ M of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ ions needs to be determined. The solution needs to be classified as acidic, basic or neutral.

Concept Introduction:

A base is defined as the substance that can give ${\text{[OH}}^{-}\text{]}$ in its solution whereas an acid can give ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ in its solution. General equation for acid and base can be written as:

  ${\text{HA}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ A}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

Or,

  ${\text{BOH}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{ + B}}^{\text{+}}{}_{\text{(aq)}}$

The ionic product of water can be used to calculate the concentration of ${\text{[OH}}^{-}\text{]}$ and ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ ions in the given solution.

  ${\text{ K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$

Experimentally proved that at $\text{25}°\text{C}$ , the concentration of ${\text{OH}}^{\text{-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is same and that is $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-7}}$ . Thus, the value of ${\text{K}}_{\text{w}}$ stands for ion product constant for water with value of $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\text{ at 25}°\text{C}$ . If the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions is more than ${\text{OH}}^{\text{-}}$ ions it is said to be acidic whereas the solution with more ${\text{OH}}^{\text{-}}$ ions than ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions it is said to be basic solution.

The acidic and basic nature of the solution can be expressed in terms of pH of the solution. The range of pH lies from 0 to 14. The pH value from 0 to 7 indicates an acidic solution whereas the pH value from 7 to 14 stands for a basic solution. A neutral solution can be expressed as pH 7. It can be calculated with the help of concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions and can be formulated as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 9.81P

  ${\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] =1}{\text{.9 × 10}}^{\text{-12}}\text{ M}$

  $\text{pH = 2}\text{.28}$

Since the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Explanation of Solution

Given Information:

  ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ ions = $\text{5}\text{.3 }\times {\text{10}}^{\text{-3}}$ M

Calculation:

Calculate ${\text{[OH}}^{\text{-}}]$ with the help of ${\text{K}}_{\text{w}}$ expression:

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ \text{ 1}{\text{.0 × 10}}^{\text{-14}}{\text{= [OH}}^{\text{-}}{}_{\text{(aq)}}\text{] [5}\text{.3 }\times {\text{10}}^{\text{-3}}\text{] }\\ {\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] =}\frac{\text{1}{\text{.0 × 10}}^{\text{-14}}}{\text{5}\text{.3 }\times {\text{10}}^{\text{-3}}}\\ {\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] =1}{\text{.9 × 10}}^{\text{-12}}\text{ M}\end{array}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Calculate $\text{pH}$ with the help of $\text{pH}$ expression:

  $\begin{array}{l}{\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH = - log [5}\text{.3 }\times {\text{10}}^{\text{-3}}\text{]}\\ \text{pH = 2}\text{.28}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The missing value of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ and $\text{pH}$ for the given solution with $\text{2}\text{.0 }\times {\text{10}}^{\text{-8}}$ M of ${\text{[OH}}^{-}\text{]}$ ions needs to be determined. The given solution needs to be classified as acidic, basic or neutral.

Concept Introduction:

A base is defined as the substance that can give ${\text{[OH}}^{-}\text{]}$ in its solution whereas an acid can give ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ in its solution. General equation for acid and base can be written as:

  ${\text{HA}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ A}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

  ${\text{BOH}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{ + B}}^{\text{+}}{}_{\text{(aq)}}$

The ionic product of water can be used to calculate the concentration of ${\text{[OH}}^{-}\text{]}$ and ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ ions in the given solution.

  ${\text{ K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$

Experimentally proved that at $\text{25}°\text{C}$ , the concentration of ${\text{OH}}^{\text{-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is same and that is $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-7}}$ . Thus, the value of ${\text{K}}_{\text{w}}$ stands for ion product constant for water with value of $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\text{ at 25}°\text{C}$ . If the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions is more than ${\text{OH}}^{\text{-}}$ ions it is said to be acidic whereas the solution with more ${\text{OH}}^{\text{-}}$ ions than ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions it is said to be basic solution.

The acidic and basic nature of the solution can be expressed in terms of pH of the solution. The range of pH lies from 0 to 14. The pH value from 0 to 7 indicates an acidic solution whereas the pH value from 7 to 14 stands for a basic solution. A neutral solution can be expressed as pH 7. It can be calculated with the help of concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions and can be formulated as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 9.81P

  ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] =5}{\text{.0× 10}}^{\text{-7}}\text{ M}$

  $\text{pH = 6}\text{.30}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Explanation of Solution

Given Information:

  ${\text{[OH}}^{\text{-}}]$ ions = $\text{2}\text{.0 }\times {\text{10}}^{\text{-8}}$ M

Calculation:

Calculate ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ with the help of ${\text{K}}_{\text{w}}$ expression:

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ \text{ 1}{\text{.0 × 10}}^{\text{-14}}\text{= [2}\text{.0 }\times {\text{10}}^{\text{-8}}{\text{] [H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] =}\frac{\text{1}{\text{.0 × 10}}^{\text{-14}}}{\text{2}\text{.0 }\times {\text{10}}^{\text{-8}}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] =5}{\text{.0× 10}}^{\text{-7}}\text{ M}\end{array}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Calculate $\text{pH}$ with the help of $\text{pH}$ expression:

  $\begin{array}{l}{\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH = - log [ 5}{\text{.0× 10}}^{\text{-7}}\text{ M]}\\ \text{pH = 6}\text{.30}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The missing value of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ and ${\text{[OH}}^{-}\text{]}$ for the given solution with $\text{pH}$ value 4.4 needs to be calculated. The given solution needs to be classified as acidic, basic or neutral.

Concept Introduction:

A base is defined as the substance that can give ${\text{[OH}}^{-}\text{]}$ in its solution whereas an acid can give ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ in its solution. General equation for acid and base can be written as:

  ${\text{HA}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ A}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

  ${\text{BOH}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{ + B}}^{\text{+}}{}_{\text{(aq)}}$

The ionic product of water can be used to calculate the concentration of ${\text{[OH}}^{-}\text{]}$ and ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ ions in the given solution.

  ${\text{ K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$

Experimentally proved that at $\text{25}°\text{C}$ , the concentration of ${\text{OH}}^{\text{-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is same and that is $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-7}}$ . Thus, the value of ${\text{K}}_{\text{w}}$ stands for ion product constant for water with value of $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\text{ at 25}°\text{C}$ . If the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions is more than ${\text{OH}}^{\text{-}}$ ions it is said to be acidic whereas the solution with more ${\text{OH}}^{\text{-}}$ ions than ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions it is said to be basic solution.

The acidic and basic nature of the solution can be expressed in terms of $\text{pH}$ of the solution. The range of pH lies from 0 to 14. The $\text{pH}$ value from 0 to 7 indicates an acidic solution whereas the $\text{pH}$ value from 7 to 14 stands for a basic solution. A neutral solution can be expressed as $\text{pH}$ 7. It can be calculated with the help of concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions and can be formulated as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 9.81P

  ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{= 4 }\times {\text{ 10}}^{\text{-5}}$

  ${\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] = 2}{\text{.5 × 10}}^{\text{-12}}\text{ M}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Explanation of Solution

Given Information:

  $\text{pH = 4}\text{.4}$

Calculation:

Calculate ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ with the help of $\text{pH}$ expression:

  $\begin{array}{l}{\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{4}{\text{.4 = - log [ H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{= 4 }\times {\text{ 10}}^{\text{-5}}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}$ with the help of ${\text{K}}_{\text{w}}$ expression:

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ \text{ 1}{\text{.0 × 10}}^{\text{-14}}{\text{= [OH}}^{\text{-}}{}_{\text{(aq)}}\text{] [4 }\times {\text{ 10}}^{\text{-5}}\text{] }\\ {\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] =}\frac{\text{1}{\text{.0 × 10}}^{\text{-14}}}{\text{4 }\times {\text{ 10}}^{\text{-5}}}\\ {\text{[OH}}^{\text{-}}{}_{\text{(aq)}}\text{] = 2}{\text{.5 × 10}}^{\text{-12}}\text{ M}\end{array}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Interpretation Introduction

(d)

Interpretation:

The missing value of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ and $\text{pH}$ for the given solution with $\text{6}\text{.8 }\times {\text{10}}^{\text{-10}}$ M of ${\text{[OH}}^{-}\text{]}$ ions needs to be calculated. The solution should be classified as acidic, basic or neutral.Concept Introduction:

A base is defined as the substance that can give ${\text{[OH}}^{-}\text{]}$ in its solution whereas an acid can give ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ in its solution. General equation for acid and base can be written as:

  ${\text{HA}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ A}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

  ${\text{BOH}}_{\text{(l)}}\underset{}{\rightleftarrows }{\text{ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{ + B}}^{\text{+}}{}_{\text{(aq)}}$

The ionic product of water can be used to calculate the concentration of ${\text{[OH}}^{-}\text{]}$ and ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ ions in the given solution.

  ${\text{ K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$

Experimentally proved that at $\text{25}°\text{C}$ , the concentration of ${\text{OH}}^{\text{-}}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is same and that is $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-7}}$ . Thus, the value of ${\text{K}}_{\text{w}}$ stands for ion product constant for water with value of $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\text{ at 25}°\text{C}$ . If the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions is more than ${\text{OH}}^{\text{-}}$ ions it is said to be acidic whereas the solution with more ${\text{OH}}^{\text{-}}$ ions than ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions it is said to be basic solution.

The acidic and basic nature of the solution can be expressed in terms of pH of the solution. The range of pH lies from 0 to 14. The pH value from 0 to 7 indicates an acidic solution whereas the pH value from 7 to 14 stands for a basic solution. A neutral solution can be expressed as pH 7. It can be calculated with the help of concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions and can be formulated as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 9.81P

  ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] = 1}{\text{.5× 10}}^{\text{-5}}\text{ M}$

  $\text{pH = 4}\text{.82}$

Since the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Explanation of Solution

Given Information:

  ${\text{[OH}}^{\text{-}}]$ ions = $\text{6}\text{.8 }\times {\text{10}}^{\text{-10}}$ M

Calculation:

Calculate ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ with the help of ${\text{K}}_{\text{w}}$ expression:

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [OH}}^{\text{-}}{}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ \text{ 1}{\text{.0 × 10}}^{\text{-14}}\text{= [6}\text{.8 }\times {\text{10}}^{\text{-10}}{\text{] [H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] =}\frac{\text{1}{\text{.0 × 10}}^{\text{-14}}}{\text{6}\text{.8 }\times {\text{10}}^{\text{-10}}}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] = 1}{\text{.5× 10}}^{\text{-5}}\text{ M}\end{array}$

Since, the concentration of is ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{] }$ more than ${\text{[OH}}^{\text{-}}]$ , therefore, the solution must be acidic.

Calculate $\text{pH}$ with the help of $\text{pH}$ expression:

  $\begin{array}{l}{\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH = - log [ 1}{\text{.5× 10}}^{\text{-5}}\text{ M]}\\ \text{pH = 4}\text{.82}\end{array}$