Question

How many milliliters of \(0.50 \mathrm{N}\) NaOH are required to neutralize exactly \(15.0 \mathrm{mL}\) of \(0.35 \mathrm{N} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short answer

21 mL of \(0.50 \mathrm{N}\) NaOH solution is required to neutralize exactly 15.0 mL of \(0.35 \mathrm{N}\) H2SO4 solution.

Step-by-step solution

Calculate the moles of H2SO4 present in the given volume

To calculate the moles of H2SO4 present in the given volume, use the formula: moles = volume × concentration (in terms of normality). moles of H2SO4 = volume of H2SO4 × concentration of H2SO4 moles of H2SO4 = \(15.0 \mathrm{mL} \times 0.35 \mathrm{N}\) First, convert the volume from mL to L: 15.0 mL = 15.0/1000 L = 0.015 L Now, plug in the values: moles of H2SO4 = \(0.015 \mathrm{L} \times 0.35 \mathrm{N}\) moles of H2SO4 = \(0.00525 \mathrm{mol}\)

Calculate the moles of NaOH required to neutralize the H2SO4

The balanced chemical equation for the reaction between NaOH and H2SO4 is: 2 NaOH + H2SO4 → Na2SO4 + 2 H2O From the balanced equation, we can see that 2 moles of NaOH are required to completely neutralize 1 mole of H2SO4. So, we will multiply the moles of H2SO4 by 2 to get the moles of NaOH required for neutralization. moles of NaOH = 2 × moles of H2SO4 moles of NaOH = 2 × \(0.00525 \mathrm{mol}\) moles of NaOH = \(0.0105 \mathrm{mol}\)

Calculate the volume of NaOH solution required

Now that we have the moles of NaOH required for neutralization, we can find the volume of NaOH solution needed using its concentration. Use the formula: volume = moles ÷ concentration. volume of NaOH = moles of NaOH ÷ concentration of NaOH volume of NaOH = \(0.0105 \mathrm{mol} \div 0.50 \mathrm{N}\) Now, calculate the volume: volume of NaOH = \(0.021 \mathrm{L}\) Convert the volume from L to mL: 0.021 L = 21 mL So, 21 mL of 0.50 N NaOH solution is required to neutralize exactly 15.0 mL of 0.35 N H2SO4 solution.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that talks about the proportions of reactants and products in chemical reactions. It involves using a balanced chemical equation to calculate the amount of substances consumed and produced. In this context, stoichiometry allows us to determine how much of one chemical is needed to react completely with a certain amount of another.

For example, in the neutralization reaction involving NaOH and H\(_2\)SO\(_4\), stoichiometry helps us understand that two moles of NaOH are needed to fully react with one mole of H\(_2\)SO\(_4\). This ratio is derived from the balanced chemical equation.

• Balanced Equation: \(2 \text{ NaOH} + \text{ H}_2\text{SO}_4 \rightarrow \text{ Na}_2\text{SO}_4 + 2\text{ H}_2\text{O}\)

Recognizing these relationships is essential for calculating correct amounts of reactants required or products formed during reactions.

Normality

Normality is a way of expressing the concentration of a solution. It is the gram equivalent weight of a solute per liter of solution. In the context of acid-base chemistry, normality can be very useful because it considers the reactive capacity of the molecules involved.

To find the normality of an acid or base, you often multiply the molarity by the number of protons an acid can donate or a base can accept. For example, sulfuric acid (H\(_2\)SO\(_4\)) is a diprotic acid, meaning each molecule can donate two protons. This is why it’s important in equations involving neutralization, where balance is key.

• The given normality of an H\(_2\)SO\(_4\) solution in this exercise is 0.35 N, indicating how concentrated the acid is in terms of its potential to react.
• Similarly, NaOH has a normality of 0.50 N, showing its capacity to neutralize an acid.

This understanding helps in calculating the volumes required for complete reactions in solutions.

Chemical Equation Balancing

Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the equation. This is crucial because it reflects the law of conservation of mass, stating that matter cannot be created or destroyed in a chemical reaction.

In our exercise, the chemical equation for the reaction between NaOH and H\(_2\)SO\(_4\) must be correctly balanced:

• \(2 \text{ NaOH} + \text{ H}_2\text{SO}_4 \rightarrow \text{ Na}_2\text{SO}_4 + 2\text{ H}_2\text{O} \)

Here, the equation shows that two molecules of NaOH are required to completely react with one molecule of H\(_2\)SO\(_4\). Balancing is fundamental for accurately understanding the stoichiometric relationships and performing further calculations in chemistry.

Acid-Base Chemistry

Acid-base chemistry focuses on reactions between acids and bases, the products of which are typically a salt and water. Neutralization is a key type of reaction in this field, where hydrogen ions from the acid combine with hydroxide ions from the base, forming water.

Let's consider sulfuric acid (H\(_2\)SO\(_4\)) and sodium hydroxide (NaOH), two common substances in acid-base chemistry. When they react, they undergo a neutralization reaction:

• 2 NaOH (aq) + H\(_2\)SO\(_4\) (aq) → Na\(_2\)SO\(_4\) (aq) + 2 H\(_2\)O (l)

Sulfuric acid, a strong diprotic acid, provides two hydrogen ions per molecule, making it a powerful participant in these reactions. Sodium hydroxide, a strong base, fully dissociates in water to give hydroxide ions. The balance of these ions determines the nature of the solution after the reaction, illustrating important principles of acid-base chemistry.