Chapter 16.6, Problem 16.12P

(a)

Interpretation Introduction

Interpretation:

Percent ionization of $0.100\text{M}$$\text{HCN}$ solution has to be determined.

Concept Introduction:

For equilibrium reaction rate of forward reaction is equal to rate of reverse direction. Consider a general reaction that is as follows:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

Expression for equilibrium constant is written as follows:

  $K_{\text{eq}}=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here,

$K_{\text{eq}}$ is equilibrium constant.

$\left[\text{A}\right]$, $\left[\text{B}\right]$, $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are concentration of substances in reaction.

$a$, $b$, $c$ and $d$ are coefficients of substances.

(b)

Interpretation Introduction

Interpretation:

Percent ionization of $0.0250\text{M}$ solution of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{5}}\text{O}$ has to be determined.

Concept Introduction:

For equilibrium reaction rate of forward reaction is equal to rate of reverse direction. Consider a general reaction that is as follows:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

Expression for equilibrium constant is written as follows:

  $K_{\text{eq}}=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here,

$K_{\text{eq}}$ is equilibrium constant.

$\left[\text{A}\right]$, $\left[\text{B}\right]$, $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are concentration of substances in reaction.

$a$, $b$, $c$ and $d$ are coefficients of substances.

Formula to calculate percent of ionization of weak acid is as follows:

  $\text{Percent}\text{ionized}=\left(\frac{\left[{\text{H}}^{+}\right]}{\left(\text{initial}\text{concetration}\right)}\right)\left(100\right)\%$