Problem 205 - Introductory Chemistry Atoms First

Question

What is the volume in milliliters of \(0.0150 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to neutralize \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(a q) ?\)

Step-by-Step Solution

Answer

66.7 mL of \(0.0150 \mathrm{M}\) NaOH solution is required to neutralize 50.0 mL of \(0.0100 \mathrm{M}\) H2SO4 solution.

Step 1: Write the balanced chemical equation

First, we need to write the balanced chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4). The balanced chemical equation is: 2 NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)

Step 2: Calculate moles of H2SO4

Next, we will calculate the moles of H2SO4 in the 50.0 mL solution. We can do this using the given concentration (0.0100 M) and the volume (50.0 mL). Remember to convert milliliters to liters: Moles of H2SO4 = (0.0100 moles/L) x (50.0 mL x (1 L / 1000 mL)) Moles of H2SO4 = 0.0005 moles

Step 3: Determine moles of NaOH required to neutralize H2SO4

Now, we will calculate the moles of NaOH needed to neutralize the moles of H2SO4 we calculated in step 2. For this, we will use the stoichiometry of the balanced chemical equation (2 moles of NaOH required for 1 mole of H2SO4): Moles of NaOH = Moles of H2SO4 x (2 moles NaOH / 1 mole H2SO4) Moles of NaOH = 0.0005 moles x (2 moles NaOH / 1 mole H2SO4) Moles of NaOH = 0.0010 moles

Step 4: Calculate the volume of NaOH solution required

Finally, we will find the volume of the 0.0150 M NaOH solution required to neutralize the H2SO4. We will use the moles of NaOH calculated in step 3 and the given concentration of the NaOH solution (0.0150 M). Remember to convert liters to milliliters: Volume of NaOH solution = (Moles of NaOH) / (Concentration of NaOH) Volume of NaOH solution = (0.0010 moles) / (0.0150 moles/L) Volume of NaOH solution = 0.0667 L Volume of NaOH solution = 0.0667 L x (1000 mL / 1 L) Volume of NaOH solution = 66.7 mL So, 66.7 mL of 0.0150 M NaOH solution is required to neutralize 50.0 mL of 0.0100 M H2SO4 solution.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is fundamental to understanding reactions like neutralization. When a reaction occurs, atoms rearrange to form new substances, but the total number of each type of atom remains constant. This means we need equations that reflect this law of conservation of mass.
For our exercise, we have a reaction between NaOH and H2SO4. To balance this equation, we start with the unbalanced equation:

• NaOH + H2SO4 → Na2SO4 + H2O

We need to ensure that there are equal numbers of each type of atom on both sides of the equation. After balancing, we get:

• 2 NaOH + H2SO4 → Na2SO4 + 2 H2O

In this balanced equation, there are 2 sodium (Na) atoms, 4 hydrogen (H) atoms, and 4 oxygen (O) atoms on each side. Balanced equations are essential because they let us perform stoichiometric calculations accurately.

Stoichiometry

Stoichiometry is the mathematical foundation that helps us relate quantities of reactants and products in a chemical reaction. It relies on the balanced chemical equation to convert between moles of different substances.
For example, from the balanced equation in our exercise:

• 2 NaOH + H2SO4 → Na2SO4 + 2 H2O

It's clear that 2 moles of NaOH react with 1 mole of H2SO4 to form products. This gives a stoichiometric ratio of 2:1 for NaOH to H2SO4.
These ratios allow scientists and students alike to calculate exactly how much of each substance is needed or produced. So, if you know the amount of one reactant, you can find the required amount of the other based on these fixed ratios.

Molarity

Molarity is a way of expressing the concentration of a solution and is essential in solving many chemistry problems, like our neutralization question. It is defined as the number of moles of solute per liter of solution, expressed as mol/L or simply M.
For instance, in our exercise:

• The molarity of H2SO4 is 0.0100 M, meaning 0.0100 moles of H2SO4 are present in every liter of solution.
• Similarly, the NaOH has a molarity of 0.0150 M.

This concept helps us determine how much solute (like NaOH or H2SO4) is in a given volume of solution. Molarity is crucial for calculating the amount of a solution needed for reactions or for preparing solutions of exact concentrations.

Volume Conversion

In chemistry, often you need to convert between different units of volume, such as milliliters and liters. This is critical for calculating the volumes of solutions required in reactions. Unlike some complex conversions, volume conversions are typically straightforward due to simple metric prefixes. For example:

• 1 liter (L) = 1000 milliliters (mL)

In our exercise problem, we start with a volume in milliliters. When working with molarity, which is in units of moles per liter, it's necessary to convert milliliters to liters:

• 50.0 mL of H2SO4 is 0.0500 L (by dividing by 1000).

Accurate volume conversion ensures precise measurements and calculations in reactions, which is why it's a valuable skill in any chemistry task.