Chapter 10, Problem 10.86EP

(a)

Interpretation Introduction

Interpretation:

For a given solution, $\left[{\text{OH}}^{\text{-}}\right]$ $\text{pH}$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and whether the solution is acidic or basic has to be determined.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

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

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

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$, more is the acidity of the solution.

Explanation of Solution

Given that,

  $\left[{\text{OH}}^{\text{-}}\right]=7.2\times {10}^{-5}$

Then,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{1×10}}^{\text{-14}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{7}{\text{.2×10}}^{\text{-5}}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-10}}\end{array}$

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (\text{1}{\text{.4×10}}^{\text{-10}})\\ \\ \text{pH}\text{=}9.86\end{array}$

In this solution $\left[{\text{OH}}^{\text{-}}\right]$ is greater than $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\text{pH}$ is more than 7.00. Hence this solution is a basic solution.

Hence this is a basic solution having $\text{pH}\text{=}9.86$ and $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-10}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

For a given solution, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ the value of $\left[{\text{OH}}^{\text{-}}\right]$, $\text{pH}$ and whether the solution is acid or basic that has to be determined.

Concept Introduction:

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is. It ranges from 0-14.  $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

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

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

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}6.3{\text{×10}}^{\text{-3}}\text{M}$

Then,

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (6.3{\text{×10}}^{\text{-3}})\\ \\ \text{pH}\text{=}2.20\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]=1\times {10}^{-14}\\ \\ \left[{\text{OH}}^{\text{-}}\right]=\frac{1\times {10}^{-14}}{6.3{\text{×10}}^{\text{-3}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}1.6{\text{×10}}^{\text{-12}}\end{array}$

In this solution $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is greater than $\left[{\text{OH}}^{\text{-}}\right]$ and $\text{pH}$ is less than 7.00 and hence the solution is acidic.

Hence this is an acidic solution having $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.6×10}}^{\text{-12}}\text{M}$ and $\text{pH}\text{=}2.20$.

(c)

Interpretation Introduction

Interpretation:

For a given $\text{pH}$, the value of the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\left[{\text{OH}}^{\text{-}}\right]$ of the solution has to be calculated along with the prediction that whether the solution is acidic or basic.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

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

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

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\text{pH}\text{=}2.00$

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ -\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=2.00\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-2.00}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.00\times {10}^{-2}\text{M}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]=1\times {10}^{-14}\\ \\ \left[{\text{OH}}^{\text{-}}\right]=\frac{1\times {10}^{-14}}{\text{1}{\text{.00×10}}^{\text{-2}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.00×10}}^{\text{-12}}\end{array}$

As in this solution $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is greater than $\left[{\text{OH}}^{\text{-}}\right]$ and $\text{pH}$ is less than 7.00 and hence the solution is acidic.

Hence this acidic solution have $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.00×10}}^{\text{-12}}\text{M}$ and $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.00\times {10}^{-2}\text{M}$.

(d)

Interpretation Introduction

Interpretation:

For a given $\text{pH}$, the value of $\left[{\text{OH}}^{\text{-}}\right]$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and whether the solution is acid or basic that has to be determined.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

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

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

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\text{pH}\text{=}12.00$

Then,

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ -\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=12.00\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-12.00}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.00\times {10}^{-12}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]=1\times {10}^{-14}\\ \\ \left[{\text{OH}}^{\text{-}}\right]=\frac{1\times {10}^{-14}}{\text{1}{\text{.00×10}}^{\text{-12}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.00×10}}^{\text{-2}}\end{array}$

In this solution $\left[{\text{OH}}^{\text{-}}\right]$ is greater than $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\text{pH}$ more than 7.00. Hence this solution is a basic solution.

Hence this basic solution have $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.00\times {10}^{-12}\text{M}$ and $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.00×10}}^{\text{-2}}\text{M}$.