Question

(a) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .\) (b) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid in a solution containing \(0.085 \mathrm{M}\) sodium butanoate.

Short answer

(a) In pure water, solving the Ka equation gives the equilibrium concentration of H+ ions as \(x = 6.1 \times 10^{-4} \mathrm{M}\). The percent ionization is \( \frac{6.1 \times 10^{-4} \mathrm{M}}{0.0075 \mathrm{M}} \times 100 \approx 8.1 \% \). (b) In the sodium butanoate solution, solving the Ka equation gives the equilibrium concentration of H+ ions as \(x = 3.08 \times 10^{-5} \mathrm{M}\). The percent ionization is \( \frac{3.08 \times 10^{-5} \mathrm{M}}{0.0075 \mathrm{M}} \times 100 \approx 0.41 \% \).

Step-by-step solution

(1) Write the chemical equilibrium

For butanoic acid, the dissociation in water can be represented by the following equation: \( HA\; \leftrightharpoons\; H^{+} + A^{-} \) Where \(HA\) stands for butanoic acid and \(A^-\) represents the conjugate base (butanoate ion).

(2) Set up an ICE table

To solve this problem, we will use an ICE (Initial, Change, Equilibrium) table to track the concentrations of each species involved in the reaction. Pure water: | | HA | H+ | A- | | --- | --- | --- | --- | | I | 0.0075 | 0 | 0 | | C | -x | +x | +x | | E | 0.0075-x | x | x | Sodium butanoate solution: | | HA | H+ | A- | | --- | --- | --- | --- | | I | 0.0075 | 0 | 0.085 | | C | -x | +x | +x | | E | 0.0075-x | x | 0.085+x |

(3) Solve for x using the Ka expression

The Ka value is given as \(1.5 \times 10^{-5}\) for butanoic acid. We will use this to solve for the equilibrium concentration of hydrogen ions (H+). For pure water: \( K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} \) \( 1.5 \times 10^{-5} = \frac{x \times x}{0.0075 - x} \) For sodium butanoate solution: \(1.5 \times 10^{-5} = \frac{x \times (0.085 + x)}{0.0075 - x}\) Solving these equations will give us the equilibrium concentration of H+ ions.

(4) Calculate percent ionization

Percent ionization is the ratio of the concentration of ionized acid to the initial concentration of the acid, multiplied by 100. We will calculate it as follows: Percent ionization in pure water: \( \frac{[H^{+}]}{[HA]} \times 100 \) Percent ionization in sodium butanoate solution: \( \frac{[H^{+}]}{[HA]} \times 100 \) Solve these expressions to find the percent ionization in both cases.

Key concepts

Chemical Equilibrium

Chemical equilibrium is the state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the rates of the forward and reverse reactions are equal. In the context of acid-base chemistry, such as with butanoic acid, this involves an acid (butanoic acid) dissociating to produce hydrogen ions and its conjugate base (butanoate ions).

Understanding chemical equilibrium is crucial to predict how a change in conditions will affect the position of equilibrium, which involves the Le Châtelier's Principle. This principle suggests that if a system at equilibrium is disturbed, the system will adjust itself to diminish the change.

In our example, the equilibrium equation is: \[ HA ightleftharpoons H^+ + A^- \]where \(HA\) represents butanoic acid, \(H^+\) is the hydrogen ion, and \(A^-\) represents the conjugate base, the butanoate ion.

This equation shows that butanoic acid partially ionizes in water, reaching a balance between the undissociated acid and its ionic products.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a useful tool in chemistry to systematically determine the changes in concentration of species in a reaction as it approaches equilibrium. It is particularly helpful in dealing with weak acid or base equilibria.

Here's how the ICE table works for our butanoic acid example:

• Initial: Begin with the initial concentrations of each species. For the butanoic acid in pure water, we start with \(0.0075\) M of \(HA\), 0 M \(H^+\), and 0 M \(A^-\).
• Change: As the reaction proceeds, changes in concentration are noted as \(-x\) for the reactant and \(+x\) for each product, since the acid ionizes to produce \(H^+\) and \(A^-\).
• Equilibrium: Calculate the final concentrations after the change has occurred and equilibrium has been established. This will be \(0.0075 - x\) for \(HA\), and \(x\) for both \(H^+\) and \(A^-\) in pure water. In the presence of sodium butanoate, these calculations adjust to account for its initial concentration.

Using the ICE table allows us to clearly visualize and calculate the equilibrium state of the system.

Ka Expression

The acid dissociation constant, \( K_a \), is a quantitative measure of the strength of an acid in solution. It is derived from the equilibrium expression for the acid dissociation reaction.

For butanoic acid, the expression using the equilibrium concentrations is:\[ K_a = \frac{[H^+][A^-]}{[HA]} \]This formula helps determine the extent to which the acid dissociates in solution. A higher \( K_a \) value means a stronger acid, which dissociates more completely. For weak acids like butanoic acid, the \( K_a \) is quite low, indicating limited ionization.

Calculating \( K_a \) involves substituting the equilibrium concentrations from the ICE table into the equation. Solving for \( x \), the concentration of ionized hydrogen ions \([H^+]\), reveals the degree of ionization under specific conditions, such as in pure water versus a buffer solution containing a conjugate base.

Conjugate Acid-Base Pairs

In acid-base chemistry, conjugate acid-base pairs consist of two species that transform into each other by the gain or loss of a proton. A conjugate pair represents the acid and its corresponding base that results from the acid losing a proton.

Regarding butanoic acid:

• When butanoic acid (\(HA\)) donates a proton to water, it forms the butanoate ion (\(A^-\)), which is its conjugate base.
• Conversely, the butanoate ion can accept a proton to re-form butanoic acid. Thus, \(HA\) and \(A^-\) together form a conjugate acid-base pair.

This concept is essential in understanding buffering systems and calculating pH changes in solutions. For instance, in solutions with added sodium butanoate, the equilibrium is influenced by the presence of the conjugate base, showing how conjugate pairs help maintain stable pH levels even when acids or bases are introduced.