Problem 61 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

The solution's pH is 8.31, and \([\text{H}_3\text{O}^+]=4.90 \times 10^{-9} \text{ M}\).

Step 1: Determine Initial Moles of Acetic Acid and NaOH

Calculate the moles of acetic acid and NaOH using the formula: moles = molarity x volume (in liters). Since both solutions are 22.0 mL or 0.022 L, and the concentrations are 0.15 M, we have:\[ n_{ ext{acetic acid}} = 0.15 imes 0.022 = 0.0033 ext{ moles} \]\[ n_{ ext{NaOH}} = 0.15 imes 0.022 = 0.0033 ext{ moles} \]

Step 2: Identify Reaction and Calculate Remaining Species

Acetic acid and NaOH react in a 1:1 ratio to form acetate ions and water:\[ \text{CH}_3\text{CO}_2\text{H} + \text{OH}^- \rightarrow \text{CH}_3\text{CO}_2^- + \text{H}_2\text{O} \]Since they are mixed in equal moles (0.0033 moles each), both acetic acid and NaOH will be completely reacted with no excess of either. This results in 0.0033 moles of acetate ions, \( \text{CH}_3\text{CO}_2^- \).

Step 3: Calculate Concentration of Acetate Ions

Calculate the total volume of the solution to find the concentration of acetate ions. Total volume is 44.0 mL (22.0 mL + 22.0 mL) or 0.044 L.\[ \text{Concentration of } \text{CH}_3\text{CO}_2^- = \frac{0.0033}{0.044} \approx 0.075 \text{ M} \]

Step 4: Determine the pH Using Acetate Ion

Use the expression for the equilibrium reaction of acetate ions with water to estimate the pH. For this calculation, the ionization of water equation applies:\[ \text{CH}_3\text{CO}_2^- + \text{H}_2\text{O} \rightleftarrows \text{CH}_3\text{CO}_2\text{H} + \text{OH}^- \]The base ionization constant \( K_b \) can be derived from:\[ K_a \times K_b = K_w \]Given \( K_a \) for acetic acid is approximately \( 1.8 \times 10^{-5} \) and \( K_w = 1.0 \times 10^{-14} \). Solve for \( K_b \):\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]Use this to find \([\text{OH}^-]\) and subsequently \(\text{pOH}\) and \(\text{pH}\). Assuming \([\text{OH}^-] = x \):\[ K_b = \frac{x^2}{0.075} \approx 5.56 \times 10^{-10} \]\[ x^2 = 5.56 \times 10^{-10} \times 0.075 \]\[ x = \sqrt{4.17 \times 10^{-11}} \approx 2.04 \times 10^{-6} \]Calculate \(\text{pOH}\) and \(\text{pH}\):\[ \text{pOH} = -\log_{10}(2.04 \times 10^{-6}) \approx 5.69 \]\[ \text{pH} = 14 - \text{pOH} = 14 - 5.69 = 8.31 \]

Step 5: Calculate Hydronium Ion Concentration

The hydronium ion concentration \([\text{H}_3\text{O}^+]\) is related to \(\text{pH}\) by:\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \]\[ [\text{H}_3\text{O}^+] = 10^{-8.31} \approx 4.90 \times 10^{-9} \text{ M} \]

Key Concepts

Hydronium Ion Concentration

The concentration of hydronium ions, \([ ext{H}_3 ext{O}^+]\), is crucial in determining the acidity or alkalinity of a solution. It's tied directly to the \( ext{pH}\), which is defined as the negative logarithm of the hydronium ion concentration. A lower \( ext{pH}\) indicates a higher concentration of hydronium ions, making the solution more acidic. Conversely, a higher \( ext{pH}\) implies a lower \([ ext{H}_3 ext{O}^+]\), suggesting an alkaline solution. To find \([ ext{H}_3 ext{O}^+]\), simply use:

• \([ ext{H}_3 ext{O}^+] = 10^{- ext{pH}}\)

As seen in the solution, the \( ext{pH}\) was calculated to be 8.31, which corresponds to \([ ext{H}_3 ext{O}^+]\) of approximately \4.90 \times 10^{-9}\ M. This means the solution is slightly basic, with fewer hydronium ions present than in a neutral solution.

Acetic Acid

Acetic acid \(\text{(CH}_3\text{CO}_2\text{H)}\) is a weak organic acid commonly known for giving vinegar its sour taste. It partially dissociates in water, meaning it does not completely break apart into ions. This is important for understanding the acid's equilibrium with its conjugate base, the acetate ion \(\text{CH}_3\text{CO}_2^-\).When mixed with bases, like \(\text{NaOH}\), acetic acid participates in a neutralization reaction. Each molecule of acetic acid reacts with one hydroxide ion (\(\text{OH}^-\)) to form an acetate ion and water:

• \(\text{CH}_3\text{CO}_2\text{H} + \text{OH}^- \rightarrow \text{CH}_3\text{CO}_2^- + \text{H}_2\text{O}\)

This reaction is crucial in pH calculations as it determines which species are present in the solution after mixing. In this exercise, the equal moles of acetic acid and \(\text{NaOH}\) resulted in their complete reaction, providing solely acetate ions in the solution.

Equilibrium Constant

In chemical reactions, the equilibrium constant \(K\) is a value that expresses the ratio of the concentration of products to reactants at equilibrium. For acids and bases, such constants are often expressed as \(K_a\) (acid dissociation constant) or \(K_b\) (base ionization constant). These constants inform us how well an acid dissociates in water or how readily a base attracts hydrogen ions to form \(\text{OH}^-\), respectively.In the case of acetic acid, its \(K_a\) is around \1.8 \times 10^{-5}\, showing it is a weak acid that doesn't dissociate fully. This \(K_a\) allows you to find the \(K_b\) for its conjugate base, the acetate ion, using the water ionization constant \(K_w\):

• \(K_a \times K_b = K_w = 1.0 \times 10^{-14}\)

From this, you can solve for \(K_b\), which is crucial in calculating the \(\text{pH}\) when acetate ions are reintroduced into solution.

Stoichiometry

Stoichiometry is about the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict how much product will form by knowing the quantities of reactants.In this exercise, stoichiometry helps determine the reaction's result when equal molarities of \(\text{CH}_3\text{CO}_2\text{H}\) and \(\text{NaOH}\) are mixed. Each reactant was present at \3.3 \times 10^{-3}\ moles, leading to their complete reaction and production of the same amount of acetate ions \(\text{CH}_3\text{CO}_2^-\):

• \(\text{CH}_3\text{CO}_2\text{H} + \text{OH}^- \rightarrow \text{CH}_3\text{CO}_2^- + \text{H}_2\text{O}\)

Thus, stoichiometry aids in calculating remaining species and concentrations, an essential step to determining the overall \(\text{pH}\) of the solution.