Chapter 16, Problem 16.119QP

Interpretation Introduction

Interpretation:

The $\text{0}\text{.80 }M$${\text{HNO}}_{\text{2}}$ solution $pH$ and parentage range has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$: 

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

  ${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

  ${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

  ${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

$pH$ Percentage Calculation: 

The percentage ionization is another method to assess acid strength

For the reaction

  $\begin{array}{l}H{A}_{\left(aq\right)}+H_2{O}_{\left(\text{l}\right)}\stackrel{}{\rightleftharpoons }{H}_3{O}_{\left(aq\right)}^{+}+A_{\left(aq\right)}^{-}\\ \%ionization=\frac{{[H^{+}]}_{eqm}}{{[HA]}_0}\times 100\end{array}$

Percent ionization relates the equilibrium $H^{+}$ concentration, ${[H^{+}]}_{eqm}$ to the initial $HA$ concentration ${[HA]}_0$