Chapter 14, Problem 14.72SP

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{250mL}$ of solution contains $\text{4}\text{.8 g}$ of $\text{LiOH}$ has to be calculated.

Concept Introduction:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH + pOH= 14}$

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

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= antilog(-pH) =10}}^{\text{-pH}}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{0}\text{.40 L}$ of solution contains $\text{0}\text{.93g}$ of $\text{HCl}$ has to be calculated.

Concept Introduction:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH + pOH= 14}$

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

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= antilog(-pH) =10}}^{\text{-pH}}$

(c)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{50}\text{.0mL}$ of $\text{0}\text{.10 M HCl}$ to a volume of $\text{1L}$ has to be calculated.

Concept Introduction:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH + pOH= 14}$

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

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= antilog(-pH) =10}}^{\text{-pH}}$

(d)

Interpretation Introduction

Interpretation:

$\text{pH}$ of mixing of $\text{100 mL}$ of $\text{2}\text{.0}\times {\text{10}}^{-3}\text{ M HCl}$ and $\text{400 mL}$ of $\text{1}\text{.0}\times {\text{10}}^{-3}{\text{ M HClO}}_4$ has to be  calculated.

Concept Introduction:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH + pOH= 14}$

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

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= antilog(-pH) =10}}^{\text{-pH}}$