Question

Calculate the hydronium ion concentration and pH of the solution that results when \(22.0 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) is mixed with \(22.0 \mathrm{mL}\) of \(0.15 \mathrm{M} \mathrm{NaOH}\).

Short answer

The hydronium ion concentration is approximately \(5.5 \times 10^{-10}\:M\) and the pH is 9.26.

Step-by-step solution

Calculate Moles of Acetic Acid and NaOH

First, determine the number of moles of acetic acid and NaOH present before the reaction. Use the formula \( ext{moles} = ext{concentration} \times ext{volume} \). For acetic acid, \( n = 0.15 ext{ M} \times 0.022 ext{ L} = 0.0033 ext{ mol} \). Similarly, for NaOH, \( n = 0.15 ext{ M} \times 0.022 ext{ L} = 0.0033 ext{ mol} \).

Determine the Reaction and Limiting Reagent

The reaction between acetic acid and NaOH is \( ext{CH}_3 ext{CO}_2 ext{H} + ext{NaOH} ightarrow ext{CH}_3 ext{CO}_2 ext{Na} + ext{H}_2 ext{O} \). Both reactants have equal moles, so they will completely react with each other leaving no excess of either reactant.

Calculate concentration of Acid-Base Conjugate

Since all acetic acid and NaOH are consumed, the product formed is sodium acetate (\( ext{CH}_3 ext{CO}_2 ext{Na} \)). The total volume of the solution is \(22.0 ext{ mL} + 22.0 ext{ mL} = 44.0 ext{ mL} = 0.044 ext{ L} \). The concentration of sodium acetate is \( [ ext{CH}_3 ext{CO}_2 ext{Na}] = \frac{0.0033 ext{ mol}}{0.044 ext{ L}} = 0.075 ext{ M} \).

Calculate Hydronium Ion Concentration

Sodium acetate in solution can undergo hydrolysis: \[ ext{CH}_3 ext{CO}_2^- + ext{H}_2 ext{O} ightleftharpoons ext{CH}_3 ext{CO}_2 ext{H} + ext{OH}^- \]. To find the concentration of hydronium ions \( [ ext{H}_3 ext{O}^+] \), we need to determine the \( ext{K}_w \) and \( ext{K}_a \) for the acetate ion: \( ext{K}_w = 1 imes 10^{-14} \) and \( ext{K}_a = 1.8 imes 10^{-5} \). The \( [ ext{H}_3 ext{O}^+] \) is given by \([ ext{H}_3 ext{O}^+] = \frac{K_w}{[ ext{OH}^-]} \), which can be calculated indirectly knowing that pH = 14 - pOH. Using the conjugate base (acetate ion) equation, \([ ext{OH}^-] = \sqrt{K_b imes [ ext{Acetate}]} = \sqrt{\frac{K_w}{K_a} \times 0.075} \), we find \([ ext{OH}^-]=1.826 imes 10^{-5} \text{ M} \).

Determine pH of the Solution

Calculate the pOH: \( ext{pOH} = - ext{log}[ ext{OH}^-] = - ext{log}(1.826 imes 10^{-5}) = 4.74 \). Using \( ext{pH} + ext{pOH} = 14 \), the pH of the solution is \( ext{pH} = 14 - 4.74 = 9.26 \). This indicates a slightly basic mixture, which is expected from the products of a neutralization reaction between a weak acid and a strong base.

Key concepts

Acid-Base Neutralization

Acid-base neutralization is a chemical reaction where an acid and a base react to form water and a salt. This process essentially neutralizes the acid and base, leading to a mixture that can be either acidic, basic, or neutral depending on the strengths and concentrations of the reactants.

In the reaction between acetic acid and sodium hydroxide, both are present in equal amounts, which means they completely react with each other, leaving no excess of these reactants behind. The balanced chemical equation for this reaction is:

• \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} + \mathrm{NaOH} \rightarrow \mathrm{CH}_3\mathrm{CO}_2\mathrm{Na} + \mathrm{H}_2\mathrm{O} \)

This results in the formation of sodium acetate and water as products.
Such reactions are key to understanding pH changes in mixtures, especially important in buffer solutions.

Acetic Acid

Acetic acid, also known as ethanoic acid, is a weak organic acid widely known as the main component of vinegar apart from water. It has the chemical formula \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \).

Being a weak acid means that it does not dissociate completely in water. This characteristic affects its reactivity and the type of products formed when it participates in neutralization reactions.
When reacting with a strong base like sodium hydroxide, acetic acid donates a proton (\( \mathrm{H}^+ \)), transforming into its conjugate base, the acetate ion (\( \mathrm{CH}_3\mathrm{CO}_2^- \)).

• This interaction shifts the pH towards the higher side, indicating a less acidic environment than before.
• The final product in this case is sodium acetate, a salt which can further affect the pH through hydrolysis processes.

Sodium Acetate

Sodium acetate (\( \mathrm{CH}_3\mathrm{CO}_2\mathrm{Na} \)) is a salt formed from the neutralization of acetic acid with sodium hydroxide. It is a noteworthy component because it plays a significant role in maintaining pH in various chemical systems.

As sodium acetate dissolves in water, it can undergo hydrolysis, a reversible interaction with water molecules that affects the pH of the solution.

This process creates a buffer system, as sodium acetate can act as a reserve of acetic acid and acetate ions. It ensures that even when acids or bases are added to the solution, the pH doesn't change drastically.

• The acetate ion from sodium acetate can also react with water to form hydroxide ions (\( \text{OH}^- \)), which raises the pH slightly above neutral.
• This kind of explanation is crucial to interpret the final observed pH of the solution.

Hydronium Ion Concentration

Hydronium ion concentration \([\text{H}_3\text{O}^+]\) is crucial to determine the acidity of a solution. In acidic reactions between a weak acid and a strong base, it helps predict the final pH.

To find \([\text{H}_3\text{O}^+]\) when sodium acetate is formed, we need to explore the relation between the equilibrium constants and the concentrations involved in the hydrolysis process:

• The dissociation constant of water, \( \text{K}_w \), is \( 1 \times 10^{-14} \).
• The acid dissociation constant, \( \text{K}_a \), for acetic acid is \( 1.8 \times 10^{-5} \).

These values help estimate \([\text{OH}^-]\) through the relation \( \sqrt{\frac{K_w}{K_a} \times \text{concentration of acetate}} \), eventually allowing us to find the pH.

Finally, from the relation that \( \text{pH} = 14 - \text{pOH} \), the slightly basic nature of the solution is confirmed, demonstrating how important understanding these calculations can be for interpreting experimental data.