Chapter 10, Problem 10.90EP

(a)

Interpretation Introduction

Interpretation:

The value $\text{pH}$ of $\text{1}\text{.0}\text{M}\text{HCl}$ is more or less than $\text{0}\text{.10}\text{M}\text{HCl}$ has to be calculated.

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

According to definition of $\text{pH}$,

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

Hence for $\text{1}\text{.0}\text{M}\text{HCl}$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log 1.0\\ \\ \text{pH}\text{=}0\end{array}$

Now, for $\text{0}\text{.10}\text{M}\text{HCl}$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log 0.10\\ \\ \text{pH}\text{=}1\end{array}$

Hence it can be concluded that $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in $\text{1}\text{.0}\text{M}\text{HCl}$ is more than in $\text{0}\text{.10}\text{M}\text{HCl}$ and so $\text{1}\text{.0}\text{M}\text{HCl}$ has lower $\text{pH}$.

Hence $\text{1}\text{.0}\text{M}\text{HCl}$ has lower $\text{pH}$ than $\text{0}\text{.10}\text{M}\text{HCl}$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}\text{.0}\text{M}\text{NaOH}$ is more or less than $\text{0}\text{.10}\text{M}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be calculated.

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

According to definition,

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

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

Now, in $\text{1}\text{.0}\text{M}\text{NaOH}$ there will be ${\text{OH}}^{-}$ and hence,

$\begin{array}{l}\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]\\ \\ \text{pOH}\text{=}-\log 1.0\\ \\ \text{pOH}\text{=}0\\ \\ \text{pH}=14-\text{pOH}\\ \\ \text{pH}=14-0\\ \\ \text{pH}=14\end{array}$

Now for $\text{0}\text{.10}\text{M}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log 0.10\\ \\ \text{pH}\text{=}1\end{array}$

Hence it can be concluded that $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in $\text{1}\text{.0}\text{M}\text{NaOH}$ is lesser than in $\text{0}\text{.10}\text{M}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and so $\text{1}\text{.0}\text{M}\text{NaOH}$ has higher $\text{pH}$.

Hence $\text{1}\text{.0}\text{M}\text{NaOH}$ has higher $\text{pH}$ than $\text{0}\text{.10}\text{M}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.10}\text{M}{\text{HClO}}_3$ is more or less than $\text{0}\text{.10}\text{M}\text{HF}$ has to be calculated.

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

${\text{HClO}}_3$ is a strong acid and it will dissociate completely in solution.

Thus $\text{pH}$ of $\text{0}\text{.10}\text{M}{\text{HClO}}_3$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log 0.10\\ \\ \text{pH}\text{=}1\end{array}$

But, $\text{HF}$ is a weak acid and it does not completely dissociate in solution and concentration of hydronium ion is less than $\text{0}\text{.10}\text{M}$.

Hence $\text{pH}$ for $\text{0}\text{.10}\text{M}\text{HF}$ will be high as it has less $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ than $\text{0}\text{.10}\text{M}{\text{HClO}}_3$.

Hence $\text{0}\text{.10}\text{M}{\text{HClO}}_3$ has lower than $\text{0}\text{.10}\text{M}\text{HF}$.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.3\times {10}^{-4}$ is more or less than solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.2\times {10}^{-4}$ has to be calculated.

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

According to definition of $\text{pH}$,

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

Hence for solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.3\times {10}^{-4}$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (7.3\times {10}^{-4})\\ \\ \text{pH}\text{=}3.14\end{array}$

Now, for solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.2\times {10}^{-4}$,

$\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (1.2\times {10}^{-4})\\ \\ \text{pH}\text{=}3.92\end{array}$

Hence it can be concluded that $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.3\times {10}^{-4}$ is more than in solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.2\times {10}^{-4}$ and so solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.3\times {10}^{-4}$ has lower $\text{pH}$.

Hence solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.3\times {10}^{-4}$ has lower $\text{pH}$ than solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.2\times {10}^{-4}$.