Chapter 18, Problem 18.173P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and ${\left[{\text{H}}_{\text{3}}\text{O}\right]}^{\text{+}}$ of the water at $\text{0}\text{C}\text{and}\text{50}{}^{\text{o}}\text{C}$ has to be calculated when ${\text{K}}_{\text{w}}\text{ = 1}{\text{.139×10}}^{\text{-15}}\text{ at 0°C and 5}{\text{.474 × 10}}^{\text{-14}}\text{ at 50°C}$.

Concept introduction:

The ${\text{K}}_w$ value is calculating by using following formula,

  ${\text{K}}_{\text{w}}\text{= }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\times \left[{\text{OH}}^{-}\right]$

The $\text{pH}$ is calculated by using following formula,

  $\text{pH}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pD}$ and ${\left[{\text{D}}_{\text{3}}\text{O}\right]}^{\text{+}}$ of the water at $\text{0}\text{C}\text{and}\text{50}{}^{\text{o}}\text{C}$ has to be calculated when ${\text{K}}_{\text{w}}\text{ = 3}{\text{.64×10}}^{\text{-16}}\text{ at 0°C and 7}{\text{.89 × 10}}^{\text{-15}}\text{ at 50°C}$.

Concept introduction:

The ${\text{K}}_w$ value is calculating by using following formula,

  ${\text{K}}_{\text{w}}\text{= }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{×}\left[{\text{OD}}^{\text{-}}\right]$

The $\text{pH}$ is calculated by using following formula,

  $\text{pH}\text{=}\text{-}\text{log}{\text{(D}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}$

(c)

Interpretation Introduction

Interpretation:

The difference between the $\text{pH}$, ${\left[{\text{H}}_{\text{3}}\text{O}\right]}^{\text{+}}$ of the water at $\text{0}\text{C}\text{and}\text{50}{}^{\text{o}}\text{C}$ and $\text{pD}$, ${\left[{\text{D}}_{\text{3}}\text{O}\right]}^{\text{+}}$ of the water at $\text{0}\text{C}\text{and}\text{50}{}^{\text{o}}\text{C}$ has to be explained.