Problem 50 - General Chemistry: Principles and Modern Applications

Question

The exact neutralization of \(10.00 \mathrm{mL}\) of \(0.1012 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text { aq })\) requires \(23.31 \mathrm{mL}\) of \(\mathrm{NaOH}\). What must be the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\)

Step-by-Step Solution

Answer

The molarity of the NaOH solution is 0.0868 M.

Step 1: Identify the Knowns

The following are identified from the problem: volume of H2SO4 solution = 10.00 mL = 0.01 L and molarity of H2SO4 = 0.1012 M; volume of NaOH = 23.31 mL = 0.02331 L. All volume values were converted from mL to L for consistency.

Step 2: Calculate moles of H2SO4

Next, calculate the number of moles of H2SO4 reacted. The mole of a substance is its molarity times its volume in litres. Consider the relation: moles = molarity x volume. So, moles of H2SO4 = 0.1012 M x 0.01 L = 0.001012 mol.

Step 3: Use stoichiometry to find moles of NaOH

In the balanced chemical equation, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, 2 times the moles of H2SO4 will give the moles of NaOH. So, moles of NaOH = 2 x moles of H2SO4 = 2 x 0.001012 mol = 0.002024 mol.

Step 4: Calculate molarity of NaOH

Molarity is defined as the number of moles per litre. The volume of NaOH solution used is 0.02331 L. Therefore, the molarity of NaOH is calculated as: Molarity of NaOH = moles of NaOH / volume of NaOH in L = 0.002024 mol / 0.02331 L = 0.0868 M

Key Concepts

Neutralization Reaction

In a neutralization reaction, an acid and a base react to form water and a salt. This type of chemical reaction is important in understanding how acids and bases interact. For example, in the given reaction, sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) reacts with sodium hydroxide (\(\mathrm{NaOH}\)). Here the neutralization produces sodium sulfate (\(\mathrm{Na}_2\mathrm{SO}_4\)) and water (\(\mathrm{H}_2\mathrm{O}\)).

• Acid: \(\mathrm{H}_2\mathrm{SO}_4\)
• Base: \(\mathrm{NaOH}\)
• Product: \(\mathrm{Na}_2\mathrm{SO}_4\) and \(\mathrm{H}_2\mathrm{O}\)

In our exercise, this reaction is used to find an unknown concentration by using the concept of stoichiometry. For each molecule of sulfuric acid, two molecules of sodium hydroxide are required to completely neutralize it. This stoichiometric ratio is crucial in molarity calculations, allowing us to solve such chemistry problems effectively.

Molarity Calculation

Molarity is a measure of concentration that describes the number of moles of a solute in one liter of solution. It's a critical concept when performing stoichiometric calculations in chemistry.

• Formula: \(\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)

To find the molarity of sodium hydroxide (\(\mathrm{NaOH}\)) in our problem, we first calculated the moles of sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) using its molarity and volume. We then used the balanced chemical equation to determine the moles of \(\mathrm{NaOH}\) needed. Following this, the molarity of \(\mathrm{NaOH}\) was calculated by dividing the mole value obtained by the volume of \(\mathrm{NaOH}\) in liters.
This step-by-step approach ensures accuracy and clarity, helping to resolve complex reactions effectively.

Balanced Chemical Equation

A balanced chemical equation is fundamental to performing stoichiometric calculations as it represents the conservation of mass in chemical reactions. It provides the necessary mole ratio used in calculating reactants and products.
The equation from the exercise is:\[\mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) + 2\, \mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + 2\, \mathrm{H}_2\mathrm{O}(1)\]This equation is balanced because it has

• 2 molecules of sodium (\(\mathrm{Na}\))
• 2 oxygen (\(\mathrm{O}\)) atoms from hydroxide (\(\mathrm{OH}^−\))
• 2 water molecules (\(\mathrm{H}_2\mathrm{O}\))

on both sides. A balanced equation not only confirms that the reaction follows the law of conservation of mass (total number of atoms remain the same) but also is crucial for calculating unknowns like molarity, using stoichiometry.Understanding how to balance and interpret these equations is key to mastering chemistry concepts.