Chapter 18, Problem 18.178P

(a)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated.

Concept Introduction:

Water dissociates very slightly to form ions and this equilibrium process is known as auto ionization or self-ionization. The auto ionization equation can be written as follows,

${\text{2 H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + OH}}^{\text{-}}$

The equilibrium constant K for this reaction can be written as follows,

$\text{K = }\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{{\left[\text{H2O}\right]}^2}$

The concentration of water is constant and can be considered as pure liquid. Hence concentration term for water can be eliminated from the expression by taking its concentration as unity. So the new constant ion-product constant for water can be expressed as follows,

$\text{K ×}{\left(\text{1}\right)}^{\text{2}}{\text{= K}}_{\text{w}}\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{(at 25}}^{\text{o}}\text{C)}$

Where,

    $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{OH}}^{\text{-}}\right]\text{ = }\left(\text{1}{\text{.0×10}}^{\text{-14}}\right)\text{=1}{\text{.0×10}}^{\text{-7}}\text{ M }{\text{(at 25}}^{\text{o}}\text{C) }$

pH of a solution can be defined as the negative of common logarithm of $\left[{\text{H}}^{\text{+}}\right]$ or  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right].$

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

(b)

Interpretation Introduction

Interpretation:

Number of  ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ required for each ions of ${\text{OH}}^{\text{-}}$ at $\text{pH 4}$ has to be calculated.

Concept Introduction:

Water dissociates very slightly to form ions and this equilibrium process is known as auto ionization or self-ionization. The auto ionization equation can be written as follows,

${\text{2 H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + OH}}^{\text{-}}$

The equilibrium constant K for this reaction can be written as follows,

$\text{K = }\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{{\left[\text{H2O}\right]}^2}$

The concentration of water is constant and can be considered as pure liquid. Hence concentration term for water can be eliminated from the expression by taking its concentration as unity. So the new constant ion-product constant for water can be expressed as follows,

$\text{K ×}{\left(\text{1}\right)}^{\text{2}}{\text{= K}}_{\text{w}}\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{(at 25}}^{\text{o}}\text{C)}$

Where,

    $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\left[{\text{OH}}^{\text{-}}\right]\text{ = }\left(\text{1}{\text{.0×10}}^{\text{-14}}\right)\text{=1}{\text{.0×10}}^{\text{-7}}\text{ M }{\text{(at 25}}^{\text{o}}\text{C) }$

pH of a solution can be defined as the negative of common logarithm of $\left[{\text{H}}^{\text{+}}\right]$ or  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right].$

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