Problem 82 - Chemistry and Chemical Reactivity

Question

For each of the following cases, decide whether the pH is less than \(7,\) equal to \(7,\) or greater than 7. (a) Equal volumes of 0.10 M acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) and \(0.10 \mathrm{M} \mathrm{KOH}\) are mixed. (b) \(25 \mathrm{mL}\) of \(0.015 \mathrm{M} \mathrm{NH}_{3}\) is mixed with \(12 \mathrm{mL}\) of 0.015 M HCl. (c) 150 mL of \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(75 \mathrm{mL}\) of 0.40 M NaOH. (d) \(25 \mathrm{mL}\) of \(0.45 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is mixed with \(25 \mathrm{mL}\) of 0.90 M NaOH.

Step-by-Step Solution

Answer

(a) pH > 7, (b) pH > 7, (c) pH = 7, (d) pH = 7.

Step 1: Analyze case (a)

We are mixing equal volumes of 0.10 M acetic acid with 0.10 M KOH. Acetic acid is a weak acid and KOH is a strong base. When mixed in equal molar amounts, the strong base will neutralize the weak acid, and we are left with the acetate ion in solution. The resulting solution's pH is governed by the weakly basic nature of the acetate ion, making the pH slightly greater than 7.

Step 2: Analyze case (b)

We have 25 mL of 0.015 M NH₃ and 12 mL of 0.015 M HCl. First, calculate the moles: Moles of NH₃ = 25 mL × 0.015 M = 0.000375 moles, and moles of HCl = 12 mL × 0.015 M = 0.00018 moles. Since HCl is a strong acid, it will react completely with NH₃, which is a weak base. However, NH₃ is in excess (0.000375 - 0.00018 = 0.000195 moles). The remaining NH₃ makes the solution basic, so the pH is greater than 7.

Step 3: Analyze case (c)

Mixing 150 mL of 0.20 M HNO₃ (strong acid) with 75 mL of 0.40 M NaOH (strong base). Calculate the moles: Moles of HNO₃ = 0.15 L × 0.20 M = 0.03 moles, and moles of NaOH = 0.075 L × 0.40 M = 0.03 moles. Since the number of moles of HNO₃ and NaOH are equal, they completely neutralize each other, resulting in a neutral solution with pH equal to 7.

Step 4: Analyze case (d)

Here, 25 mL of 0.45 M H₂SO₄ is mixed with 25 mL of 0.90 M NaOH. H₂SO₄ is a strong diprotic acid. First, calculate the moles: Moles of H₂SO₄ = 0.025 L × 0.45 M × 2 (for diprotic) = 0.0225 moles, and moles of NaOH = 0.025 L × 0.90 M = 0.0225 moles. Since the acid and the base moles are equal, they neutralize each other completely, resulting in a neutral solution with pH equal to 7.

Key Concepts

Strong Acids and Bases

Strong acids and bases play a pivotal role in chemical reactions and pH calculations due to their complete ionization in aqueous solutions. A strong acid, like hydrochloric acid (HCl), dissociates completely into hydrogen ions (H⁺) and chloride ions (Cl⁻).
This complete ionization means that the concentration of H⁺ ions in the solution is equal to the initial concentration of the acid.
Similarly, strong bases such as potassium hydroxide (KOH) dissociate completely to produce hydroxide ions (OH⁻).

• Strong acids have low pH values, often close to zero.
• Strong bases have high pH values, typically around 14.
• Mixing equal moles of a strong acid and a strong base results in neutralization, with a pH close to 7, as seen in the reaction between NaOH and HNO₃ in case (c) of the exercise.

Understanding the nature of strong acids and bases helps predict the outcome of neutralization reactions and their impact on pH levels.

Weak Acid and Strong Base Reactions

When a weak acid reacts with a strong base, the weak acid does not completely dissociate in the solution. Take acetic acid ( CH₃COOH) for example, which only partially ionizes in water.
Combining a weak acid with a strong base results in the production of a conjugate base, making the solution slightly basic.
This principle is illustrated in case (a) where acetic acid reacts with KOH.
The reaction leaves behind acetate ions, which makes the solution slightly basic:

• Weak acid: Partially dissociates, less H⁺ released.
• Strong base: Fully dissociates, more OH⁻ released.
• Resulting pH: Greater than 7, solutions are mildly basic.

This nuanced behavior is essential for understanding reactions involving weak acids and interpreting pH changes resulting from their interactions with strong bases.

Moles and Molarity Calculations

Moles and molarity are essential for understanding chemical concentrations and reactions. Moles measure the quantity of substance, while molarity denotes moles per liter of solution.
To find moles, multiply the volume of the solution by the molarity.
In case (b), for instance, we calculate:For NH₃: \( ext{moles} = 25 ext{ mL} \times 0.015 ext{ M} = 0.000375 ext{ moles} \)
For HCl: \( ext{moles} = 12 ext{ mL} \times 0.015 ext{ M} = 0.00018 ext{ moles} \)

• Understanding moles helps determine which reactant is in excess.
• Molarity calculations aid in understanding concentration changes post-reaction.

Accurate moles and molarity calculations are fundamental for predicting the outcome of reactions and determining if a solution will be acidic, neutral, or basic.

Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. These reactions often lead to significant changes in pH, particularly when strong acids and bases are involved.
For example, in case (d), an equal number of moles of sulfuric acid and sodium hydroxide neutralize each other:\[ ext{H}_2 ext{SO}_4 + 2 ext{NaOH} \rightarrow ext{Na}_2 ext{SO}_4 + 2 ext{H}_2 ext{O} \]In this case:

• Each diprotic acid (H₂SO₄) molecule requires two molecules of NaOH.
• Neutralization results in a neutral solution with a pH of 7 after reaction.
• Completion of the reaction can be checked through the moles relationship.

Neutralization is crucial for understanding how different combinations of acids and bases affect solution pH, and comprehending these principles helps solve many chemistry problems effectively.