Question

What volume of \(1.00 \mathrm{M} \mathrm{NaOH}\) is required to neutralize each of the following solutions? a. \(25.0 \mathrm{~mL}\) of \(0.154 \mathrm{M}\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(35.0 \mathrm{~mL}\) of \(0.102 \mathrm{M}\) hydrofluoric acid, \(\mathrm{HF}\) c. \(10.0 \mathrm{~mL}\) of \(0.143 \mathrm{M}\) phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(35.0 \mathrm{~mL}\) of \(0.220 \mathrm{M}\) sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Short answer

a. $3.85 \mathrm{~mL}$ of $1.00 \mathrm{M} \mathrm{NaOH}$ is required to neutralize the acetic acid solution. b. $3.57 \mathrm{~mL}$ of $1.00 \mathrm{M} \mathrm{NaOH}$ is required to neutralize the hydrofluoric acid solution. c. For phosphoric acid, the balanced equation is $3\mathrm{NaOH} + \mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{Na}_{3}\mathrm{PO}_{4} + 3\mathrm{H}_{2}\mathrm{O}$. d. For sulfuric acid, the balanced equation is $2\mathrm{NaOH} + \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + 2\mathrm{H}_{2}\mathrm{O}$.

Step-by-step solution

a. Neutralizing Acetic Acid

First, determine the moles of acetic acid (HC2H3O2) present in the solution: moles of HC2H3O2 = volume of HC2H3O2 × concentration of HC2H3O2 moles of HC2H3O2 = \(25.0 mL × 0.154 \frac{moles}{L}\) Note: Convert volume in mL to L. moles of HC2H3O2 = \(\frac{25.0}{1000} L × 0.154 \frac{moles}{L}= 0.00385 moles \) Now, use the stoichiometry of the neutralization reaction to convert moles of HC2H3O2 to moles of NaOH: HC2H3O2 + NaOH → NaC2H3O2 + H2O From the balanced equation, we can see that 1 mole of HC2H3O2 reacts with 1 mole of NaOH. moles of NaOH = moles of HC2H3O2 moles of NaOH = 0.00385 moles Finally, use the molarity of NaOH solution to find the volume of NaOH required for the neutralization: volume of NaOH = \(\frac{moles~of~NaOH}{concentration~of~NaOH}\) volume of NaOH = \(\frac{0.00385~moles}{1.00~\frac{moles}{L}} = 0.00385 L\) Note: Convert volume in L to mL. volume of NaOH = \(0.00385 L × \frac{1000}{1} mL = 3.85 mL \) Therefore, 3.85 mL of 1.00 M NaOH is required to neutralize 25.0 mL of 0.154 M acetic acid solution.

b. Neutralizing Hydrofluoric Acid

Repeat the above steps for hydrofluoric acid (HF) solution: 1. Determine the moles of HF present in the solution: moles of HF = \(35.0 mL × 0.102 \frac{moles}{L}\) moles of HF = \(\frac{35.0}{1000} L × 0.102 \frac{moles}{L}= 0.00357 moles \) 2. Determine the moles of NaOH needed for the neutralization using the balanced reaction: HF + NaOH → NaF + H2O moles of NaOH = moles of HF moles of NaOH = 0.00357 moles 3. Calculate the volume of NaOH needed: volume of NaOH = \(\frac{0.00357~moles}{1.00~\frac{moles}{L}} = 0.00357 L\) volume of NaOH = \(0.00357 L × \frac{1000}{1} mL = 3.57 mL \) Hence, 3.57 mL of 1.00 M NaOH is required to neutralize 35.0 mL of 0.102 M hydrofluoric acid solution. All similar steps can be applied for phosphoric acid (H3PO4) and sulfuric acid (H2SO4). In case of phosphoric acid, it has three reactive hydrogen atoms and in case of sulfuric acid, it has two reactive hydrogen atoms. So, the balanced equations will be: 3NaOH + H3PO4 → Na3PO4 + 3H2O 2NaOH + H2SO4 → Na2SO4 + 2H2O Keep this in mind while calculating the moles of NaOH required in part c and d.

Key concepts

Stoichiometry

Stoichiometry is at the heart of chemistry as it deals with the quantitative relationships between the reactants and products involved in chemical reactions. By understanding stoichiometry, one is equipped to predict the amounts of products that can be formed from a given amount of reactants, or conversely, the amounts of reactants needed to produce a desired quantity of product.

For example, in the neutralization of acetic acid ( HC2H3O2) with sodium hydroxide (NaOH), the stoichiometry of the reaction tells us that one mole of acid reacts with one mole of base to produce water and sodium acetate ( NaC2H3O2). This straightforward 1:1 ratio simplifies calculations for this particular reaction and is fundamental in predicting how much NaOH is needed to neutralize a known volume and concentration of acetic acid.

To improve students' understanding, it's helpful to illustrate the use of stoichiometry in step-by-step calculations, like converting the volume of acid to moles using molarity, then using the reaction's stoichiometry to find the moles of base required, and finally, determining the volume of base needed using its molarity. Including visual aids or interactive components could further enhance comprehension.

Molarity

Molarity, commonly represented as M, is a measure of the concentration of a solution and is defined as the number of moles of solute per liter of solution. It allows chemists to quantify the amount of a substance dissolved in a unit volume of liquid, making it an invaluable tool for preparing solutions and performing calculations in reactions.

For the neutralization reaction described in our exercise, the molarity of the acetic acid (0.154 M) and the sodium hydroxide (1.00 M) play a pivotal role. By using the molarity and given volume of acetic acid, you can calculate the moles of acetic acid present, which, through stoichiometry, gives you the corresponding moles of NaOH needed since the molar ratio is 1:1 for this reaction.

Clarifying the concept of molarity through examples and its application in calculations, such as determining the required volume of a 1.00 M NaOH solution to neutralize a given amount of acid, can assist students in mastering this concept. Visual representations of molarity, like the dissolving process of a substance in water, may also act as an effective educational aid.

Acid-Base Neutralization

Acid-base neutralization is a type of chemical reaction where an acid and a base react to form water and a salt. It's one of the classical reactions taught in chemistry due to its simplicity and the clear-cut concept that acids donate protons (H+) and bases accept them. In the classroom, these reactions are not only used to discuss the reaction mechanisms but also as practical examples to apply stoichiometry and molarity concepts.

Using the example from the original exercise, acetic acid ( HC2H3O2) neutralizes with sodium hydroxide (NaOH) to form sodium acetate and water. Each molecule of acetic acid reacts with a molecule of sodium hydroxide, demonstrating a simple 1:1 ratio. However, this ratio can differ, such as with sulfuric acid (H2SO4), which has two protons to donate, requiring two moles of NaOH for full neutralization.

For educational reinforcement, emphasizing the role of balanced chemical equations in determining the reaction stoichiometry is key. Additionally, students could benefit from practice exercises that involve predicting the products of neutralization reactions and calculating the required reagents' volumes and molarities to achieve neutralization. Through repeated application and illustration of these principles, students can gain a solid understanding of acid-base neutralization reactions.