Question

Given \(\mathrm{K}_{\mathrm{i}}\) for acetic acid is \(1.8 \times 10^{-5}\), calculate the percentage of ionization of \(0.5 \mathrm{M}\) acetic acid. The dissociation reaction is $$ \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \rightleftarrows \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} $$

Short answer

The percentage of ionization of 0.5 M acetic acid is approximately 0.6%.

Step-by-step solution

Write the equilibrium expression using the dissociation constant Ki

The equilibrium expression for the given dissociation reaction is: $$K_i = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]}$$ We will use this expression along with the given value of Ki to solve for the concentration of dissociated ions.

Define the initial and equilibrium concentrations

Let x be the concentration of dissociated ions, i.e. $$[\mathrm{H}^{+}] = [\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}]$$ Initially, $$[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}] = 0.5M$$ At equilibrium, $$[\mathrm{H}^{+}] = [\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}] = x$$ and $$[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}] = 0.5 - x$$

Substitute the equilibrium concentrations into the equilibrium expression and solve for x

Using the given value for Ki, plug in the equilibrium concentrations into the equilibrium expression: $$1.8\times10^{-5} = \frac{x^2}{0.5-x}$$ We can assume that x is much smaller than 0.5 since the ionization is likely to be very small. This simplification allows us to rewrite the equation as: $$1.8\times10^{-5} \approx \frac{x^2}{0.5}$$ Now, solve for x: $$x^2 = 1.8\times10^{-5} \times 0.5$$ $$x = \sqrt{9\times10^{-6}} \approx 3\times10^{-3}$$

Calculate the percentage of ionization

To find the percentage of ionization, divide the concentration of dissociated ions (x) by the initial concentration of acetic acid (0.5 M), and multiply by 100: $$\% \: \text{ionization} = \frac{3\times10^{-3}}{0.5} \times 100 \approx 0.6\%$$ The percentage of ionization of 0.5 M acetic acid is approximately 0.6%.

Key concepts

Dissociation Constant

The dissociation constant, often denoted as \( K_i \) or \( K_a \), is a measure of the extent to which a compound dissociates into its ions in a given solvent—usually water. For weak acids like acetic acid, this constant is quite small, indicating that the acid does not dissociate completely. Instead, it establishes an equilibrium between the undissociated acid and the ion constituents.

In the case of acetic acid (\( \text{HC}_2\text{H}_3\text{O}_2 \)), the dissociation constant \( K_i \) is given as \( 1.8 \times 10^{-5} \). This value signifies that only a small fraction of the acetic acid molecules donate protons to form hydronium and acetate ions.

• Use the \( K_i \) value to predict the behavior of the acid in solution.
• Compare with other acids to determine relative strengths.
• Help in calculating the concentrations of ions at equilibrium.

Equilibrium Expression

In a chemical equilibrium, the equilibrium expression is a mathematical representation that relates the concentrations of the products and reactants. For a weak acid like acetic acid, the equilibrium expression is derived from its dissociation reaction.

The reaction for acetic acid is: \[ \text{HC}_2\text{H}_3\text{O}_2 \rightleftarrows \text{H}^+ + \text{C}_2\text{H}_3\text{O}_2^- \] The equilibrium expression for this reaction is: \[ K_i = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \] This equation is crucial in predicting how the concentrations of acetic acid and its ions change at equilibrium.

Filling in the values of the concentrations can solve for unknowns and give us a quantitative understanding of the solution's composition at equilibrium.

Percentage Ionization

Percentage ionization provides insight into the extent of dissociation that takes place in a solution of a weak acid or base. It is calculated by dividing the concentration of ions formed by the initial concentration of the acid, and then multiplying by 100 to get a percentage.

For acetic acid, after finding \( x \), the concentration of \([\text{H}^+]\) at equilibrium, the percentage ionization is determined using: \[ \% \: \text{ionization} = \frac{x}{0.5} \times 100 \] With \( x \approx 3 \times 10^{-3} \), the ionization of acetic acid is approximately 0.6%.

• This small percentage highlights why acetic acid is considered a weak acid.
• Percentage ionization helps in understanding the acid's behavior in various solutions.
• Important in calculating the acid's pH and its effects in mixtures.

Equilibrium Concentrations

Equilibrium concentrations are vital in understanding the balance of products and reactants in a chemical reaction at equilibrium. For acetic acid, initial and equilibrium concentrations must be calculated to understand its degree of dissociation.

The initial concentration of acetic acid is given as 0.5 M, with the dissociated parts at \( x \). At equilibrium, the concentrations are calculated as: - \( [\text{H}^+] = x \) - \( [\text{C}_2\text{H}_3\text{O}_2^-] = x \) - \( [\text{HC}_2\text{H}_3\text{O}_2] = 0.5 - x \) These relations allow for calculations using the equilibrium expression, simplifying complex reactions into tangible numbers.