Question

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with \(0.100 M\) KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

Short answer

The pH of the solution after the addition of various volumes of KOH is as follows: (a) 20.0 mL: pH = 1.78 (b) 24.9 mL: pH = 3.54 (c) 25.0 mL: pH = 7 (d) 25.1 mL: pH = 10.46 (e) 30.0 mL: pH = 12.1

Step-by-step solution

(a) 20.0 mL KOH

Before the equivalence point, we have an excess of acid in the solution. First, we will calculate the moles of remaining \(\mathrm{HNO}_{3}\) and the new total volume of the solution. Moles of \(\mathrm{HNO}_{3}\) = (0.010 L)(0.250 mol/L) = 0.00250 mol Moles of KOH = (0.020 L)(0.100 mol/L) = 0.00200 mol Since we have an excess of \(\mathrm{HNO}_{3}\), we will subtract the moles of KOH from the moles of \(\mathrm{HNO}_{3}\). Moles of remaining \(\mathrm{HNO}_{3}\) = 0.00250 mol - 0.00200 mol = 0.00050 mol Now, we calculate the new total volume of the solution (acid + base): Total Volume = 10.0 mL + 20.0 mL = 30.0 mL or 0.030 L Next, we calculate the concentration of remaining \(\mathrm{HNO}_{3}\): Concentration = 0.00050 mol / 0.030 L = 0.0167 M Since \(\mathrm{HNO}_{3}\) is a strong acid, it completely dissociates in water. The pH can be calculated using the remaining concentration of \(\mathrm{HNO}_{3}\): \( pH = -\log_{10} [H^{+}] \) \( pH = -\log_{10} [0.0167] \) pH = 1.78 Hence, the pH after adding 20.0 mL of KOH solution is 1.78.

(b) 24.9 mL KOH

First, we calculate the moles of added KOH: Moles of KOH = (0.0249 L)(0.100 mol/L) = 0.00249 mol This volume of KOH is just before the equivalence point, so we still have an excess of \(\mathrm{HNO}_{3}\) in the solution. We subtract the moles of KOH from the moles of \(\mathrm{HNO}_{3}\): Moles of remaining \(\mathrm{HNO}_{3}\) = 0.00250 mol - 0.00249 mol = 0.00001 mol Now, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 24.9 mL = 34.9 mL or 0.0349 L Next, we calculate the concentration of remaining \(\mathrm{HNO}_{3}\): Concentration = 0.00001 mol / 0.0349 L = 0.00029 M Finally, we calculate the pH using the concentration of \(\mathrm{HNO}_{3}\): \( pH = -\log_{10} [0.00029] \) pH = 3.54 Hence, the pH after adding 24.9 mL of KOH solution is 3.54.

(c) 25.0 mL KOH

At the equivalence point, the moles of acid and base are equal. In this case, moles of KOH = 0.00250 mol, which means this is equal to the moles of \(\mathrm{HNO}_{3}\). Since nitric acid and potassium hydroxide are strong acid and strong base, they react to form a neutral solution. Total Volume = 10.0 mL + 25.0 mL = 35.0 mL or 0.035 L pH = 7 Hence, the pH at the equivalence point (after adding 25.0 mL of KOH) is 7.

(d) 25.1 mL KOH

Now, we are in the region past the equivalence point, where we have an excess of base in the solution. We calculate the moles of remaining KOH: Moles of added KOH = (0.0251 L)(0.100 mol/L) = 0.00251 mol Moles of remaining KOH = 0.00251 mol - 0.00250 mol = 0.00001 mol Next, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 25.1 mL = 35.1 mL or 0.0351 L Then, we calculate the concentration of remaining KOH: Concentration = 0.00001 mol / 0.0351 L = 0.00029 M Now, we calculate the pOH using the concentration of remaining KOH: \( pOH = -\log_{10} [OH^{-}] \) \( pOH = -\log_{10} [0.00029] \) pOH = 3.54 Next, we calculate the pH using the relationship: pH = 14 - pOH pH = 14 - 3.54 pH = 10.46 Hence, the pH after adding 25.1 mL of KOH solution is 10.46.

(e) 30.0 mL KOH

We are still in the region past the equivalence point. We calculate the moles of remaining KOH: Moles of added KOH = (0.030 L)(0.100 mol/L) = 0.00300 mol Moles of remaining KOH = 0.00300 mol - 0.00250 mol = 0.00050 mol Next, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 30.0 mL = 40.0 mL or 0.040 L Then, we calculate the concentration of remaining KOH: Concentration = 0.00050 mol / 0.040 L = 0.0125 M Now, we calculate the pOH using the concentration of remaining KOH: \( pOH = -\log_{10} [0.0125] \) pOH = 1.90 Next, we calculate the pH using the relationship: pH = 14 - pOH pH = 14 - 1.90 pH = 12.1 Hence, the pH after adding 30.0 mL of KOH solution is 12.1.

Key concepts

pH Calculation

Calculating the pH of a solution during an acid-base titration is a crucial skill. pH is a measure of the acidity or basicity of a solution, calculated as the negative logarithm (base 10) of the hydrogen ion concentration \( [H^+] \). When nitric acid (\( \text{HNO}_3 \)) reacts with potassium hydroxide (\( \text{KOH} \)), a neutralization reaction occurs, creating water and a salt. During the titration, steps involve determining the concentration of leftover acid or base in the solution.

For example, before reaching the equivalence point, there is still excess acid like in case (a) or (b). After completing the neutralization at the equivalence point, you calculate pH based on remaining constituents, potentially either HNO3 or OH- from KOH in steps that follow beyond the equivalence point. Using formulas like \( pH = -\log_{10} [H^+] \) for acidic solutions or \( pH = 14 - \text{pOH} \) for basic solutions, we derive accurate pH values. Each scenario has slight calculation differences but the underlying principle of finding the concentration of remaining ions and applying logarithms to translate these concentrations into pH remains consistent.

Equivalence Point

When performing a titration, the equivalence point is a critical juncture where the amount of acid equals the amount of base added. At this stage, a complete neutralization reaction occurs, forming water and a salt. During the titration of a strong acid like \( \text{HNO}_3 \) with a strong base like \( \text{KOH} \), the pH at the equivalence point is typically neutral, around 7.

The equivalence point is not necessarily characterized by a visible change in the solution but rather by a change in its pH. Here, precise measurements of volume are essential to identifying this point in a titration experiment. In this context, step (c) precisely illustrates that when 25.0 mL of \( \text{KOH} \) is added, the moles of base exactly match the moles of acid, resulting in a neutral solution with pH 7.

Understanding the equivalence point aids in predicting the complete neutralization of reactants and importantly, indicates how close the process is to finishing. After reaching the equivalence point, any added base will shift the solution's pH to the basic side.

Neutralization Reaction

During an acid-base titration, a neutralization reaction comes into play to balance the acidic and basic components of a chemical solution. A neutralization reaction typically involves an acid reacting with a base to produce water and a salt. This process is essential as it underpins the concept of titration, particularly between strong acids and bases like nitric acid and potassium hydroxide.

Consider the chemical reaction: \[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \]. This reaction illustrates how the hydrogen ions ( ext{H}^+) from the acid combine with hydroxide ions ( ext{OH}^-) from the base, resulting in water (essentially neutral).

The success of a titration relies on accurately determining the point at which neutralization completes, which is the equivalence point. Precise measuring and understanding the stoichiometric relationships in reactions ensure a successful titration. This means knowing how acids and bases will interact and what the expected products will be after neutralization - crucial for chemists and students alike.