Chapter 6, Problem 6.40P

Interpretation Introduction

Interpretation:

The concentration of the ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$ be at equilibrium in a 1.000L solution of 0.100 mol phenol, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$, is to be determined. The percent dissociation of the acid is to be determined.

Concept introduction:

When an acid is dissolved in water, it donates a proton to water, producing hydronium ions and its conjugate base. In most cases, only a fraction of the dissolved acid dissociates (donates a proton to water), and an equilibrium between the products and undissociated acid is set up. The concentrations of the undissociated acid and the products depend on the strength of the acid, and are determined by the acid dissociation constant Ka. The difference between the initial concentration of the acid, and the concentration of at equilibrium divided by the initial concentration is fractional dissociation, which when multiplied by 100 gives the percent dissociation.

Answer to Problem 6.40P

The concentration of the ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$ be at equilibrium is $\text{3}{\text{.16×10}}^{\text{-3}}\text{ M}$.

The acid percent dissociation is $\text{0}\text{.00316 \%}$.

Explanation of Solution

The initial concentration of phenol is calculated as

${\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH] = }\frac{\text{moles of phenol}}{\text{solution volume (L)}}\text{ = }\frac{\text{0}\text{.100 mol}}{\text{1}\text{.000 L}}\text{ = 0}\text{.100M}$

Phenol dissociation in aqueous solution can be represented as:

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$

The dissociation constant of phenol is $\text{1}{\text{.0 × 10}}^{\text{-10}}$

The equilibrium concentration is calculated by setting up a chart listing the concentrations of the acid and dissociation products as below. The fraction of the initial concentration that dissociates, equal to the difference in initial and equilibrium concentrations, is assumed to be ‘x’.

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$	${\text{H}}_{\text{2}}\text{O}$	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial	$\text{0}\text{.1 M}$		$\text{0}$	$\text{0}$
Change	$\text{-x}$		$\text{+x}$	$\text{+x}$
Equilibrium	$\text{(0}\text{.1- x)}$		$\text{+x}$	$\text{+x}$

The acid dissociation constant expression can then be written in terms of the equilibrium concentrations from the chart. Substituting the values of initial concentration and the acid dissociation constant, the expression is solved for the value of ‘x’.

$\begin{array}{l}\text{Ka = }\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]}\\ \text{1}{\text{.0×10}}^{\text{-10}}\text{= }\frac{\text{(x)(x)}}{\text{(0}\text{.100-x)}}\\ \text{x = 3}{\text{.16×10}}^{\text{-6}}\text{ M}\end{array}$

‘x’ is also the concentration of hydronium ions and the conjugate base ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$

Therefore, the concentration of the ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$ at equilibrium is $\text{3}{\text{.16×10}}^{\text{-3}}\text{ M}$

The formula to calculate the percent dissociation as follows:

$\begin{array}{l}\text{Percent dissociation =}\frac{{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]}_{\text{eq}}}{{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{O}H\right]}_{\text{initial}}}\text{×100}\\ \text{Percent dissociation =}\frac{\text{ 3}{\text{.16×10}}^{\text{-6}}\text{ M}}{\text{0}\text{.100 M}}\text{×100}\\ =\text{3}{\text{.16×10}}^{\text{-3}}\text{ M}\end{array}$

Therefore, the percent dissociation is $\text{3}{\text{.16 ×10}}^{\text{-3}}\text{ \%}$

Conclusion

The concentration of the ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$ at equilibrium and acid percent dissociation is calculated using the value of its acid dissociation constant and initial concentration.