Problem 48 - Chemistry: Principles and Reactions

Question

A \(0.1375 \mathrm{M}\) solution of potassium hydroxide is used to titrate \(35.00 \mathrm{~mL}\) of \(0.257 M\) hydrobromic acid. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of potassium hydroxide is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{KOH}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

Step-by-Step Solution

Answer

Question: Write the balanced net ionic equation for the reaction between potassium hydroxide (KOH) and hydrobromic acid (HBr), and state the species present at the equivalence point. Calculate the volume of KOH required to reach the equivalence point, and determine the pH of the solution before any KOH is added, halfway to the equivalence point, and at the equivalence point. Answer: The balanced net ionic equation is OH⁻(aq) + H⁺(aq) → H2O(l). At the equivalence point, K⁺(aq), Br⁻(aq), and H2O(l) are present. The volume of KOH required is 64.73 mL. The pH before adding any KOH is 0.59, halfway to the equivalence point is 1.29, and at the equivalence point is 7.

Step 1: (a) Balanced net ionic equation

The balanced net ionic equation for the reaction between potassium hydroxide and hydrobromic acid is given by: OH⁻(aq) + H⁺(aq) → H2O(l)

Step 2: (b) Species present at the equivalence point

At the equivalence point, all of the hydrobromic acid (HBr) has reacted with the potassium hydroxide (KOH) and their respective cations and anions are present in the solution. Therefore, the species present at the equivalence point are K⁺(aq), Br⁻(aq), and H2O(l).

Step 3: (c) Volume of potassium hydroxide

To find the volume of potassium hydroxide (KOH) required to reach the equivalence point, we can use the equation: \(n_{O H^{-}}=n_{H^+}\) moles of OH⁻ = moles of H⁺ \(V_{KOH}×C_{KOH}=V_{HBr}×C_{HBr}\) Where \(V_{KOH}\) is the volume of KOH, and \(C_{KOH}\), \(V_{HBr}\), and \(C_{HBr}\) are known. Plugin the values: \(V_{KOH} = \frac{V_{HBr} * C_{HBr}}{C_{KOH}} = \frac{35.00 \mathrm{~mL} * 0.257 \mathrm{M}}{0.1375 \mathrm{M}}\) \(V_{KOH} = 64.73 \mathrm{~mL}\)

Step 4: (d) pH before any KOH is added

The pH of the solution before adding any KOH is the pH of the hydrobromic acid (HBr) solution. Since HBr is a strong acid, it will dissociate completely in water: \([\mathrm{H^+}] = 0.257 \mathrm{M}\) pH \(= -\log([\mathrm{H^+}])\) pH \(= -\log(0.257) \approx 0.59\)

Step 5: (e) pH halfway to the equivalence point

Halfway to the equivalence point, half of the strong acid has been neutralized by the strong base. The resulting solution is a mixture of the strong acid and its conjugate weak base. At this point: \([\mathrm{H^+}] = \frac{0.257 \,\mathrm{M}}{2}\) pH \(= -\log([\mathrm{H^+}])\) pH \(= -\log(\frac{0.257}{2}) \approx 1.29\)

Step 6: (f) pH at the equivalence point

At the equivalence point, all of the strong acid has reacted with the strong base. Since both are strong, the resulting solution is neutral (pH = 7), and any remaining ions do not affect the pH. pH at the equivalence point \(= 7\).

Key Concepts

Net Ionic Equation

In an acid-base titration, the net ionic equation simplifies the chemical reaction to show only the species that undergo a change. When potassium hydroxide (KOH) reacts with hydrobromic acid (HBr), we focus on the exchange of ions. KOH dissociates in water to form K⁺ and OH⁻ ions, while HBr dissociates into H⁺ and Br⁻ ions.
The reaction between the hydroxide ions (OH⁻) from KOH and the hydrogen ions (H⁺) from HBr forms water (H₂O).
This can be represented by the net ionic equation:

• OH⁻(aq) + H⁺(aq) → H₂O(l)

Neutralization takes place as the acid and base react to form water, highlighting the essential components of the acid-base reaction.

Equivalence Point

The equivalence point in a titration marks the moment when equal numbers of moles of acid and base have reacted. At this juncture, all the HBr has reacted with KOH, meaning every H⁺ ion has found an OH⁻ ion. Importantly, the solution now contains neither excess acid nor excess base.
What remains in the solution are:

• Potassium ions (K⁺)
• Bromide ions (Br⁻)
• Water (H₂O)

These leftover ions do not react further, as they are spectator ions. The equivalence point of a strong acid and strong base reaction leads to a neutral pH, typically around 7, due to the absence of strong acidic or basic characters among the products.

pH Calculation

The pH of a solution in a titration changes as one moves through the following stages: beginning, halfway to the equivalence point, and at equivalence point.
**Before Any KOH is Added:**

- The pH of HBr (being a strong acid) is given by the concentration of H⁺ ions. - Here, \[ ext{pH} = -\log([0.257]) \approx 0.59 \\]**Halfway to Equivalence Point:**

- Half of the acid is neutralized, the solution consists of partially neutralized acid leading to: \[\text{pH} = -\log([\frac{0.257}{2}]) \approx 1.29\]**At Equivalence Point:**

- Complete neutralization results in a pH of 7. - This indicates a neutral solution, characteristic of reactions between strong acids and bases as no excess hydronium or hydroxide ions remain.

Molarity and Volume

Molarity (M) measures the concentration of a solution in terms of moles of solute per liter of solution, while volume is typically measured in milliliters (mL) or liters in a titration.
To determine the volume of KOH needed to reach the equivalence point:
- Use the relationship: \[ \text{moles of } OH⁻ = \text{moles of } H⁺ \] - Given the molarities and the volume of acid, this relationship can be rewritten as:

• \( V_{KOH} \times C_{KOH} = V_{HBr} \times C_{HBr} \)
• \( V_{KOH} = \frac{V_{HBr} \times C_{HBr}}{C_{KOH}} \)

- Plugging in the values,\[ V_{KOH} = \frac{35.00 \mathrm{~mL} \times 0.257 \mathrm{M}}{0.1375 \mathrm{M}} \approx 64.73 \mathrm{~mL}\]This calculation enables us to predict and measure the exact volume required to achieve equivalence, ensuring all the acid is perfectly neutralized by the base.