Problem 41 - Chemistry: The Central Science in SI Units, Global Edition

Question

How many milliliters of \(0.0750 \mathrm{M} \mathrm{KOH}\) are required to titrate each of the following solutions to the equivalence point: \((\mathbf{a}) 30.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HCOOH},\) (b) \(45.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{HNO}_{3},\) (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(3.00 \mathrm{~g}\) of \(\mathrm{HBr}\) per liter?

Step-by-Step Solution

Answer

To titrate the given solutions to the equivalence point, the required volumes of 0.0750 M KOH are as follows: (a) 36.0 mL for 30.0 mL of 0.0900 M HCOOH (b) 45.0 mL for 45.0 mL of 0.0750 M HNO3 (c) 24.7 mL for 50.0 mL of a solution containing 3.00 g of HBr per liter.

Step 1: Problem (a)

Determine the volume of 0.0750 M KOH solution required to titrate 30.0 mL of 0.0900 M HCOOH solution. 1. Write the balanced chemical equation: \(HCOOH + KOH \rightarrow KCOOH + H_2O\) 2. Calculate the moles of \(HCOOH\): moles = concentration × volume moles = \(0.0900 M \times 0.0300\: L\) moles = \(2.70 \times 10^{-3} mol\) 3. Determine the mole ratio between \(HCOOH\) and \(KOH\): From the balanced chemical equation, for each mole of \(HCOOH\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 4. Calculate the volume of the KOH solution required: moles of KOH = moles of \(HCOOH\) moles of KOH = \(2.70 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{2.70 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0360\: L\) or \(36.0\: mL\)

Step 2: Problem (b)

Determine the volume of 0.0750 M KOH solution required to titrate 45.0 mL of 0.0750 M HNO3 solution. 1. Write the balanced chemical equation: \(HNO_{3} + KOH \rightarrow KNO_{3} + H_{2}O\) 2. Calculate the moles of \(HNO_{3}\): moles = concentration × volume moles = \(0.0750 M \times 0.0450\: L\) moles = \(3.375 \times 10^{-3} mol\) 3. Determine the mole ratio between \(HNO_{3}\) and \(KOH\): From the balanced chemical equation, for each mole of \(HNO_{3}\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 4. Calculate the volume of the KOH solution required: moles of KOH = moles of \(HNO_{3}\) moles of KOH = \(3.375 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{3.375 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0450\: L\) or \(45.0\: mL\)

Step 3: Problem (c)

Determine the volume of 0.0750 M KOH solution required to titrate 50.0 mL of a solution containing 3.00 g of HBr per liter. 1. Write the balanced chemical equation: \(HBr + KOH \rightarrow KBr + H_2O\) 2. Calculate the concentration of the HBr solution: Concentration = \(\frac{3.00\: g/L}{80.91\: g/mol}\) (molar mass of HBr = 80.91 g/mol) Concentration = \(0.0371\: M\) 3. Calculate the moles of \(HBr\): moles = concentration × volume moles = \(0.0371\: M \times 0.0500\: L\) moles = \(1.855 \times 10^{-3} mol\) 4. Determine the mole ratio between \(HBr\) and \(KOH\): From the balanced chemical equation, for each mole of \(HBr\), there is one mole of \(KOH\). Therefore, the mole ratio is 1:1. 5. Calculate the volume of the KOH solution required: moles of KOH = moles of HBr moles of KOH = \(1.855 \times 10^{-3} mol\) Volume of KOH solution = moles of KOH / concentration of KOH Volume = \(\frac{1.855 \times 10^{-3} mol}{0.0750 M}\) Volume = \(0.0247\: L\) or \(24.7\: mL\)

Key Concepts

Balanced Chemical Equation

In an acid-base titration, the balanced chemical equation is a vital tool. It allows us to predict the products of the reaction and understand the stoichiometry involved. The central idea is that acids react with bases to neutralize each other. In these reactions, acids donate protons (H\(^+\)), which are accepted by bases. For instance:

• Formic acid (\(HCOOH\)) reacts with potassium hydroxide (\(KOH\)) to form potassium formate (\(KCOOH\)) and water: \(HCOOH + KOH \rightarrow KCOOH + H_{2}O\).
• Nitric acid (\(HNO_{3}\)) neutralizes \(KOH\) forming potassium nitrate (\(KNO_{3}\)) and water: \(HNO_{3} + KOH \rightarrow KNO_{3} + H_{2}O\).
• Hydrobromic acid (\(HBr\)) reacts similarly: \(HBr + KOH \rightarrow KBr + H_{2}O\).

The balanced equation shows us that, in each case, one mole of acid reacts with one mole of base. Understanding these reactions helps in determining the exact amount of base needed during the titration process.

Mole Concept

The mole concept is a fundamental principle in chemistry. It connects the mass of a substance to the number of particles it contains. When dealing with titrations, knowing the moles of reactants helps us understand how much substance is needed to achieve a reaction. For example:

• The moles of formic acid (\(HCOOH\)) are calculated as: \(0.0900\, M \times 0.0300\, L = 2.70 \times 10^{-3}\, mol\).
• For nitric acid \(HNO_{3}\), moles are: \(0.0750\, M \times 0.0450\, L = 3.375 \times 10^{-3}\, mol\).
• The moles of hydrobromic acid \(HBr\) are derived from its concentration: \(0.0371\, M \times 0.0500\, L = 1.855 \times 10^{-3}\, mol\).

Understanding moles helps us relate how much of one substance will react with another. In our titration examples, the 1:1 mole ratio from the balanced equations tells us that the moles of acid will equal the moles of base required, guiding precise calculations.

Molarity Calculation

Molarity is a measure of how concentrated a substance is in a solution. It is expressed as moles of solute per liter of solution (mol/L). Calculating molarity is essential when preparing solutions for titrations.To find molarity, divide the number of moles of the solute by the volume of the solution in liters. This is crucial in acid-base titrations, where it helps determine the volume of titrant needed for neutralization. For instance:

• In the case of \(HBr\), given 3.00 g/L with molar mass 80.91 g/mol, molarity = \(\frac{3.00\, g/L}{80.91\, g/mol} = 0.0371\, M\).

By understanding molarity, one can accurately calculate how much titrant is required to reach the equivalence point, ensuring an effective and precise titration.

Equivalence Point

The equivalence point in a titration is where the acid and base exactly neutralize one another. At this point, the number of moles of acid equals the number of moles of base. This is a crucial stage in a titration, as it signals that the reaction is complete.During our titration examples, each reaction reached its equivalence point when moles of \(KOH\) equaled the moles of the respective acids (\(HCOOH\), \(HNO_{3}\), \(HBr\)). To determine how much \(KOH\) was required, we used the mole relations provided by the balanced equations. For each mole of acid present, one mole of \(KOH\) was used.Finding the equivalence point allows us to precisely measure the concentration of an unknown solution or to confirm the expected concentration. It forms the basis of accurate and reliable titration results.