Question

(a) Calculate the percent ionization of \(0.0085 \mathrm{Mbutanoic}\) acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.0085 \mathrm{M}\) butanoic acid in a solution containing \(0.075 M\) sodium butanoate.

Short answer

In the absence of sodium butanoate, the percent ionization of 0.0085 M butanoic acid is approximately 1.41%. However, in the presence of 0.075 M sodium butanoate, the percent ionization of 0.0085 M butanoic acid significantly decreases, to approximately 0.02%.

Step-by-step solution

Write the Ka expression

The Ka expression for butanoic acid (\(CH_{3}CH_{2}CH_{2}COOH\)) dissociating into protons and butanoate (\(CH_{3}CH_{2}CH_{2}COO^-\)) can be written as follows: \[K_a = \frac{[H^+][CH_{3}CH_{2}CH_{2}COO^{-}]}{[CH_{3}CH_{2}CH_{2}COOH]}\]

Set up the ICE table

To find the concentration of the ionized species, we can use the Initial, Change, and Equilibrium (ICE) table method. Initially, we have 0.0085 M butanoic acid, and no dissociation. So the initial concentrations are: [H+] = 0 [Butanoate] = 0 [Butanoic acid] = 0.0085 M As the acid dissociates, we get the following changes in concentrations: [H+] = +x [Butanoate] = +x [Butanoic acid] = -x At equilibrium, the concentrations are: [H+] = x [Butanoate] = x [Butanoic acid] = 0.0085 - x

Calculate the proton concentration ([H+])

We can now substitute these values into the Ka expression: \[1.5\times10^{-5} = \frac{x\cdot x}{0.0085 - x}\] It is reasonable to assume that x << 0.0085, so we can simplify the equation as follows: \[1.5\times10^{-5} \approx \frac{x^2}{0.0085}\] Now we can solve for x (the proton concentration): \[x = \sqrt{1.5\times10^{-5} \times 0.0085} \approx 1.20\times10^{-4} M\]

Calculate the percent ionization

Now that we have the proton concentration, we can find the percent ionization of butanoic acid by dividing the proton concentration by the initial concentration of butanoic acid and multiplying by 100: Percent ionization = \(\frac{[H^+]}{[Initial\:butanoic\:acid]}\) × 100 Percent ionization = \(\frac{1.20\times10^{-4}}{0.0085}\) × 100 ≈ 1.41 % b) Percent ionization of 0.0085 M butanoic acid in a 0.075 M sodium butanoate solution

Consider the effect of sodium butanoate

Sodium butanoate (\(CH_{3}CH_{2}CH_{2}COONa\)) is the salt of butanoic acid, which means it can affect the ionization of butanoic acid. Its presence results in a common ion effect, as sodium butanoate dissociates into butanoate and sodium ions when dissolved in water. This will change the equilibrium concentration of the butanoate ion, and thus affect the percent ionization. Initial concentration of butanoate ions = 0.075 M. The new ICE table Equilibrium concentrations are: [H+] = x [Butanoate] = 0.075 + x [Butanoic acid] = 0.0085 - x

Calculate the new proton concentration ([H+])

Using the new equilibrium concentrations, we can rewrite the Ka expression: \[1.5\times10^{-5} = \frac{x(0.075 + x)}{0.0085 - x}\] We can again assume that x << 0.0085 and x << 0.075. Therefore, we can simplify the equation as follows: \[1.5\times10^{-5} \approx \frac{x\cdot 0.075}{0.0085}\] Now solving for x: \[x \approx \frac{1.5\times10^{-5} \times 0.0085}{0.075} \approx 1.70\times10^{-6} M\]

Calculate the new percent ionization

Now that we have the new proton concentration, we can find the percent ionization of butanoic acid in the presence of sodium butanoate: Percent ionization = \(\frac{[H^+]}{[Initial\:butanoic\:acid]}\) × 100 Percent ionization = \(\frac{1.70\times10^{-6}}{0.0085}\) × 100 ≈ 0.02 % In conclusion, the percent ionization of 0.0085 M butanoic acid is: a) ≈ 1.41 % in the absence of sodium butanoate b) ≈ 0.02 % in the presence of 0.075 M sodium butanoate

Key concepts

Ionization

In chemistry, ionization refers to the process by which an atom or a molecule converts into ions by gaining or losing electrons. For acids like butanoic acid, ionization involves the acid dissociating into hydrogen ions ( [H^+] ) and the corresponding anion (butanoate). During ionization, the equilibrium for weak acids is established, meaning that not all the acid molecules will dissociate in a solution. The given equation helps to calculate how much of the acid actually turns into ions, which is crucial to understanding the acid's strength and behavior. Weak acids have incomplete ionization, meaning they do not entirely dissociate in water, and their equilibrium lies more to the left, favoring the undissociated form. The concentration of hydrogen ions ( [H^+] ) at equilibrium helps us to calculate the percent ionization and reflects the extent of ionization.

Common Ion Effect

The common ion effect is a key concept in acid-base equilibrium. It occurs when a solution contains two substances that share a common ion. This effect significantly impacts ionization and ultimately the equilibrium of acids and bases. In our problem, the common ion effect comes into play with the addition of sodium butanoate to the butanoic acid solution.

• Sodium butanoate dissociates in water to provide butanoate ions ( [CH_{3}CH_{2}CH_{2}COO^-] ), which is a common ion already present from the ionization of butanoic acid.
• This increase in concentration of the common ion shifts the equilibrium towards the left, reducing the acid's ionization.

The common ion effect makes acids appear weaker in solutions containing their corresponding salt, as observed with the significantly reduced percent ionization of butanoic acid when sodium butanoate is present.

Dissociation Constant (Ka)

The dissociation constant, (K_a), is a significant measure in understanding the strength of an acid. It quantifies the degree to which an acid dissociates into its constituent ions at equilibrium. For butanoic acid, the K_a value is given as 1.5 \times 10^{-5}, indicating it is a weak acid. The larger K_a value signifies that the acid is stronger with a greater degree of ionization.The K_a expression for an acid is derived from the concentrations of the ions at equilibrium:\[K_a = \frac{[H^+][A^-]}{[HA]}\]Where [HA], [H^+], and [A^-] represent the equilibrium concentrations of the undissociated acid, hydrogen ion, and the anion respectively.Being able to calculate and understand K_a is essential for predicting how an acid will behave in different situations, such as during the common ion effect or when determining the extent of percent ionization.

Percent Ionization

Percent ionization gives us insight into how much of an acid has dissociated in solution, as a percentage of the entire initial amount. Knowing this percentage allows chemists to determine the strength and efficiency of an acid in various conditions, which is particularly useful in chemistry experiments and industrial processes.The formula we utilize for determining percent ionization is:\[\text{Percent ionization} = \left(\frac{[H^+]}{[\text{Initial acid concentration}]}\right) \times 100\]In the provided scenario, calculating the percent ionization requires determining the concentration of [H^+] ions formed at equilibrium, comparing this to the starting concentration of the acid solution.A higher percent ionization suggests the acid is more dissociated, whereas a lower percent indicates fewer hydrogen ions in solution, often impacted by factors like concentration of common ions or presence of salts.