Problem 43 - Chemistry The Central Science In Si Units

Question

A \(10.0-\mathrm{mL}\) sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with \(0.100 \mathrm{MKOH}\) solution. Calculate the \(\mathrm{pH}\) 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) \(30.0 \mathrm{~mL}\)

Step-by-Step Solution

Answer

20.0 mL: pH 1.78, 24.9 mL: pH 3.54, 25.0 mL: pH 7.00, 25.1 mL: pH 10.45, 30.0 mL: pH 12.10.

Step 1: Understand the Initial Conditions

We start with a 10.0 mL sample of 0.250 M \( \mathrm{HNO}_{3} \), a strong acid. This gives us the initial moles of \( \mathrm{HNO}_{3} \): \( 0.010 \text{ L} \times 0.250 \dfrac{\text{mol}}{\text{L}} = 0.0025 \text{ mol} \). We will titrate this with a 0.100 M \( \mathrm{KOH} \), a strong base, meaning the reaction will go to completion. The reaction is: \( \mathrm{HNO}_{3} + \mathrm{KOH} \rightarrow \mathrm{KNO}_{3} + \mathrm{H}_2\mathrm{O} \).

Step 2: Volume Added: 20.0 mL

Calculate moles of \( \mathrm{KOH} \) added: \( 0.020 \text{ L} \times 0.100 \dfrac{\text{mol}}{\text{L}} = 0.0020 \text{ mol} \). \( \mathrm{HNO}_{3} \) leftover: \( 0.0025 - 0.0020 = 0.0005 \text{ mol} \). The total volume is 30.0 mL or 0.030 L. \( \mathrm{H}^{+} \) concentration: \( \dfrac{0.0005}{0.030} = 0.0167 \text{ M} \). pH: \( -\log(0.0167) = 1.78 \).

Step 3: Volume Added: 24.9 mL

Calculate moles of \( \mathrm{KOH} \) added: \( 0.0249 \text{ L} \times 0.100 \dfrac{\text{mol}}{\text{L}} = 0.00249 \text{ mol} \). \( \mathrm{HNO}_{3} \) leftover: \( 0.0025 - 0.00249 = 0.00001 \text{ mol} \). The total volume is 34.9 mL or 0.0349 L. \( \mathrm{H}^{+} \) concentration: \( \dfrac{0.00001}{0.0349} = 0.000287 \text{ M} \). pH: \( -\log(0.000287) = 3.54 \).

Step 4: Volume Added: 25.0 mL (Equivalence Point)

Calculate moles of \( \mathrm{KOH} \) added: \( 0.025 \text{ L} \times 0.100 \dfrac{\text{mol}}{\text{L}} = 0.0025 \text{ mol} \). Now both moles of \( \mathrm{KOH} \) and \( \mathrm{HNO}_{3} \) are equal, thus neutralizing each other. The solution will contain pure water and salt, leading to a neutral pH of 7.00.

Step 5: Volume Added: 25.1 mL

At 25.1 mL added, \( 0.0251 \text{ L} \times 0.100 \dfrac{\text{mol}}{\text{L}} = 0.00251 \text{ mol} \) of \( \mathrm{KOH} \) has been added. Excess \( \mathrm{OH}^{-} \): \( 0.00251 - 0.0025 = 0.00001 \text{ mol} \). Total volume: 35.1 mL or 0.0351 L. \( \mathrm{OH}^{-} \) concentration: \( \dfrac{0.00001}{0.0351} \approx 0.000285 \text{ M} \). pOH: \( -\log(0.000285) \approx 3.55 \). pH: \( 14 - 3.55 = 10.45 \).

Step 6: Volume Added: 30.0 mL

Calculate moles of \( \mathrm{KOH} \) added: \( 0.030 \text{ L} \times 0.100 \dfrac{\text{mol}}{\text{L}} = 0.0030 \text{ mol} \). Excess \( \mathrm{OH}^{-} \): \( 0.0030 - 0.0025 = 0.0005 \text{ mol} \). Total volume: 40 mL or 0.040 L. \( \mathrm{OH}^{-} \) concentration: \( \dfrac{0.0005}{0.040} = 0.0125 \text{ M} \). pOH: \( -\log(0.0125) = 1.90 \). pH: \( 14 - 1.90 = 12.10 \).

Key Concepts

pH calculation

The pH of a solution is a measure of the acidity or basicity of that solution. As we engage in titration, understanding the stepwise addition of a base to an acid helps us determine the pH changes.
In the provided exercise, each step calculates the moles of the base added, tracks any remaining acid, and then derives the pH:

• The initial moles of the acid help us keep track of which component is left after each addition of base.
• Once we calculate the concentration of either \([H^+]\) or \([OH^-]\), the pH or pOH can be obtained using the logarithmic formula: \(-\log([H^+])\) for pH or \(-\log([OH^-])\) for pOH.
• The relationship \(pH + pOH = 14\) allows for conversion between the two.

These calculations guide us in understanding the acid-base neutrality and beyond, capturing the dynamic changes occurring during the neutralization process.

neutralization reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This is what we're observing in the titration exercise, specifically with \(\mathrm{HNO}_3\) and \(\mathrm{KOH}\).
When these substances meet, they participate in a chemical reaction:

• The hydrogen ions \(H^+\) from the acid react with the hydroxide ions \(OH^-\) from the base.
• This exchange produces water, \(H_2O\), and a salt, \(\mathrm{KNO}_3\).
• As the reaction progresses, the remaining acid or base changes, impacting the solution's pH.

With every drop of base added, we're progressively consuming the available \(H^+\) ions in the solution, where the ultimate goal is to achieve stoichiometric equivalence, meaning equal amounts of acid and base.

equivalence point

The equivalence point is a crucial marker in a titration where the number of moles of added base equals the initial number of moles of acid in the solution.
This point signifies complete neutralization, resulting in a balanced reaction:

• In the case of our \(\mathrm{HNO}_3\) and \(\mathrm{KOH}\) titration, reaching equivalence means all the acid has been neutralized by the base.
• At the equivalence point, the solution effectively has neutral pH, assuming strong acids and bases, which in this instance resulted in a pH of 7.00.
• Understanding the equivalence point allows us to measure the progress of the titration and confirm the chemical transformation is complete.

Beyond the equivalence point, any additional base will result in an increase in pH, as seen in the steps following the equivalence calculations.