Chapter 6, Problem 6.36P

(a)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}^{\text{+}}$ and $\text{pH}$ of $\text{0}{\text{.010M HNO}}_{\text{3}}$ has to be calculated.

Concept introduction:

The $\text{pH}$ of solution can be calculated as,

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

Answer to Problem 6.36P

$\left[{\text{H}}^{\text{+}}\right]$ = $\text{0}\text{.010M}$

$\text{pH = 2}\text{.00}$

Explanation of Solution

Concentration of the given solution is $\text{0}\text{.010M}$

The concentration of hydrogen ion is equal to the actual concentration of the solution.

Therefore, $\left[{\text{H}}^{\text{+}}\right]$ = $0.010M$

$\begin{array}{l}\text{pH= -log}\left[{\text{H}}^{\text{+}}\right]\\ \\ =\text{-log(0}\text{.010)}\\ \\ =2.00\end{array}$

(b)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}^{\text{+}}$ and $\text{pH}$ of $\text{0}\text{.035M KOH}$ has to be calculated.

Concept introduction:

The $\text{pH}$ of solution can be calculated as,

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

When calculate $\left[{\text{OH}}^{\text{-}}\right]$ an ionic product of water formula has written,

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]=\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\end{array}$

Where,

${\text{K}}_{\text{w}}\text{=1}\text{.0}\times {\text{10}}^{\text{-14}}$

Answer to Problem 6.36P

$\left[{\text{OH}}^{-}\right]$ = $\text{0}\text{.035 M}$

$\text{pH = 12}\text{.54}$

Explanation of Solution

Concentration of the given solution is $\text{0}\text{.035 M}$

The concentration of hydrogen ion is equal to the actual concentration of the solution.

Therefore, $\left[{\text{OH}}^{-}\right]$ = $0.035M$

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{= }\frac{\text{1}{\text{.0×10}}^{{}^{\text{-14}}}}{\text{0}\text{.035 M}}\text{=2}{\text{.86×10}}^{\text{-13}}\text{M}\\ \\ \text{ pH= -log}\left(\text{2}{\text{.86×10}}^{\text{-13}}\right)\\ \\ =12.54\end{array}$

(c)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}^{\text{+}}$ and $\text{pH}$ of $\text{0}\text{.030M HCl}$ has to be calculated.

Concept introduction:

The $\text{pH}$ of solution can be calculated as,

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

Answer to Problem 6.36P

$\left[{\text{H}}^{\text{+}}\right]$ = $\text{0}\text{.030M}$

$\text{pH = 1}\text{.52}$

Explanation of Solution

Concentration of the given solution is $\text{0}\text{.030M}$

The concentration of hydrogen ion is equal to the actual concentration of the solution.

Therefore, $\left[{\text{H}}^{\text{+}}\right]$ = $0.030M$

$\begin{array}{l}\text{pH= -log}\left[{\text{H}}^{\text{+}}\right]\\ \\ =\text{-log(0}\text{.030)}\\ \\ =1.52\end{array}$

(d)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}^{\text{+}}$ and $\text{pH}$ of $\text{3}\text{.0M HCl}$ has to be calculated.

Concept introduction:

The $\text{pH}$ of solution can be calculated as,

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

Answer to Problem 6.36P

$\left[{\text{H}}^{\text{+}}\right]$ = $\text{3}\text{.0M}$

$\text{pH = -0}\text{.48}$

Explanation of Solution

Concentration of the given solution is $\text{3}\text{.0M}$

The concentration of hydrogen ion is equal to the actual concentration of the solution.

Therefore, $\left[{\text{H}}^{\text{+}}\right]$ = $3.0M$

$\begin{array}{l}\text{pH= -log}\left[{\text{H}}^{\text{+}}\right]\\ \\ =\text{-log(3}\text{.0)}\\ \\ =-0.48\end{array}$

(e)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}^{\text{+}}$ and $\text{pH}$ of $\text{0}\text{.010 M }\left[{{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}{\text{N}}^{\text{+}}\right]{\text{OH}}^{\text{-}}$ has to be calculated.

Concept introduction:

The $\text{pH}$ of solution can be calculated as,

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

When calculate $\left[{\text{OH}}^{\text{-}}\right]$ an ionic product of water formula has written,

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]=\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\end{array}$

Where,

${\text{K}}_{\text{w}}\text{=1}\text{.0}\times {\text{10}}^{\text{-14}}$

Answer to Problem 6.36P

$\left[{\text{OH}}^{-}\right]$ = $\text{0}\text{.010 M}$

$\text{pH = 12}\text{.00}$

Explanation of Solution

Concentration of the given solution is $\text{0}\text{.010 M}$

The concentration of hydrogen ion is equal to the actual concentration of the solution.

Therefore, $\left[{\text{OH}}^{-}\right]$ = $0.010M$

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{= }\frac{\text{1}{\text{.0×10}}^{{}^{\text{-14}}}}{\text{0}\text{.010 M}}\text{=1}{\text{.0×10}}^{\text{-12}}\text{M}\\ \\ \text{ pH= -log}\left(\text{1}{\text{.0×10}}^{\text{-12}}\right)\\ \\ =12.00\end{array}$