Problem 81 - Chemistry: The Central Science

Question

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH}\) ? (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of \(\mathrm{AgNO}_{3}\) is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of \(\mathrm{KOH}\) must be present in the solution?

Step-by-Step Solution

Answer

a) 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. b) 768.90 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2. c) The molarity of the AgNO3 solution is 0.408066 M. d) There are 0.274152 grams of KOH in the solution.

Step 1: Write the balanced chemical equation.

The balanced chemical equation for this reaction is: \( \mathrm{HClO_{4}} + \mathrm{NaOH} \rightarrow \mathrm{NaClO_{4}} + \mathrm{H_{2}O} \)

Step 2: Calculate the moles of NaOH.

Using the given volume and concentration of NaOH, we can calculate the moles of NaOH: Moles of NaOH = volume × concentration Moles of NaOH = 0.050 L × 0.0875 mol/L Moles of NaOH = 0.004375 mol

Step 3: Calculate the moles of HClO4.

Looking at the balanced equation, the moles of HClO4 and NaOH have a 1:1 ratio. Moles of HClO4 = Moles of NaOH Moles of HClO4 = 0.004375 mol

Step 4: Calculate the volume of HClO4.

Using the concentration of HClO4, we can calculate the volume by rearranging the formula for molarity: Volume of HClO4 = moles of HClO4 ÷ concentration Volume of HClO4 = 0.004375 mol ÷ 0.115 mol/L Volume of HClO4 = 0.038043 L or 38.04 mL b) Neutralizing HCl and Mg(OH)2

Step 1: Write the balanced chemical equation.

The balanced chemical equation for this reaction is: \(2 \mathrm{HCl} + \mathrm{Mg(OH)_{2}} \rightarrow \mathrm{MgCl}_{2} + 2 \mathrm{H_{2}O} \)

Step 2: Calculate the moles of Mg(OH)2.

First, we need to find the molar mass of Mg(OH)2: Molar mass of Mg(OH)2 = 1 × 24.31 (Mg) + 2 × 15.999 (O) + 2 × 1.008 (H) = 58.32 g/mol Now we can calculate the moles of Mg(OH)2: Moles of Mg(OH)2 = mass ÷ molar mass Moles of Mg(OH)2 = 2.87 g ÷ 58.32 g/mol Moles of Mg(OH)2 = 0.049209 mol

Step 3: Calculate the moles of HCl.

From the balanced equation, we can see that the ratio between HCl and Mg(OH)2 is 2:1. Therefore: Moles of HCl = 2 × moles of Mg(OH)2 Moles of HCl = 2 × 0.049209 mol Moles of HCl = 0.098418 mol

Step 4: Calculate the volume of the HCl solution.

The volume of HCl can be found using the moles of HCl and the given concentration of HCl: Volume of HCl = moles of HCl ÷ concentration Volume of HCl = 0.098418 mol ÷ 0.128 mol/L Volume of HCl = 0.768898 L or 768.90 mL c) Precipitating AgCl from KCl

Step 1: Write the balanced chemical equation.

The balanced chemical equation for this reaction is: \( \mathrm{AgNO}_{3} + \mathrm{KCl} \rightarrow \mathrm{AgCl} + \mathrm{KNO}_{3} \)

Step 2: Calculate the moles of KCl.

First, we need to find the molar mass of KCl: Molar mass of KCl = 1 × 39.10 (K) + 1 × 35.45 (Cl) = 74.55 g/mol Now we can calculate the moles of KCl: Moles of KCl = mass ÷ molar mass Moles of KCl = 0.785 g ÷ 74.55 g/mol Moles of KCl = 0.010527 mol

Step 3: Calculate the moles of AgNO3.

From the balanced equation, the moles of AgNO3 and KCl have a 1:1 ratio. Moles of AgNO3 = moles of KCl Moles of AgNO3 = 0.010527 mol

Step 4: Calculate the molarity of the AgNO3 solution.

We can find the molarity of the AgNO3 solution using the moles of AgNO3 and the given volume of the solution: Molarity of AgNO3 = moles of AgNO3 ÷ volume Molarity of AgNO3 = 0.010527 mol ÷ 0.0258 L Molarity of AgNO3 = 0.408066 M d) Neutralizing HCl and KOH

Step 1: Write the balanced chemical equation.

The balanced chemical equation for this reaction is: \( \mathrm{HCl} + \mathrm{KOH} \rightarrow \mathrm{KCl} + \mathrm{H_{2}O} \)

Step 2: Calculate the moles of HCl.

Using the given volume and concentration of HCl, we can calculate the moles of HCl: Moles of HCl = volume × concentration Moles of HCl = 0.0453 L × 0.108 mol/L Moles of HCl = 0.0048864 mol

Step 3: Calculate the moles of KOH.

Looking at the balanced equation, the moles of HCl and KOH have a 1:1 ratio. Moles of KOH = moles of HCl Moles of KOH = 0.0048864 mol

Step 4: Calculate the mass of KOH in the solution.

First, we need to find the molar mass of KOH: Molar mass of KOH = 1 × 39.10 (K) + 1 × 15.999 (O) + 1 × 1.008 (H) = 56.107 g/mol Now we can calculate the mass of KOH: Mass of KOH = moles of KOH × molar mass of KOH Mass of KOH = 0.0048864 mol × 56.107 g/mol Mass of KOH = 0.274152 g Thus, there are 0.274152 grams of KOH in the solution.

Key Concepts

Molarity

Understanding the concept of molarity is critical for successful acid-base titration calculations. Molarity, denoted as M, is a measure of concentration within a given volume of solution. It is defined as the number of moles of solute present per liter of solution. Mathematically, it's expressed as:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In titration problems, molarity allows chemists to find out how much acid is required to neutralize a base, or vice versa. When calculating, it's important to measure volumes accurately and convert to liters if necessary, as molarity is always expressed in terms of liters of solution. This concept ensures that students can approach the stoichiometry of a neutralization reaction with proper quantitative information.

Stoichiometry

Stoichiometry is at the heart of chemical reactions, providing a quantitative relationship between reactants and products. In the context of an acid-base titration, stoichiometry allows us to predict the outcome of a neutralization reaction.

Using Ratios in Equations

The balanced chemical equations provide ratios by which molecules interact. For example, in a simple acid-base reaction such as \( \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \), the ratio of acid to base is 1:1. This means one mole of HCl will neutralize one mole of NaOH. Understanding this ratio makes it possible to determine the amount of one reactant needed to completely react with a given amount of another reactant.

Applying Stoichiometry to Titration

To calculate volumes and concentrations in titration, you need to first understand the stoichiometry of the chemical equation. This informs you of the mole-to-mole relationships necessary to perform the rest of the calculations accurately. By multiplying these ratios with the determined molarity, the volume of the solution required for neutralization can be found, as demonstrated in the step-by-step solutions.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. This is a subset of double displacement reactions and is integral to the concept of titrations in chemistry.

Acid-Base Neutralization

The general form of a neutralization reaction can be written as: \( \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \). During an acid-base titration, the goal is often to figure out the unknown concentration of an acid or a base. This is achieved by adding a reactant of known concentration until the neutralization point is reached, which is usually indicated by a color change due to an indicator.
When performing calculations for these reactions, it is essential to write out the balanced chemical equation first to understand the molar relationships between the acid and the base, as seen in the step-by-step solutions. With this information, you can use stoichiometry to calculate the necessary volumes or concentrations for neutralization.

Molar Mass

Molar mass is a fundamental concept in stoichiometry and is defined as the mass of one mole of a given substance. It is typically expressed in units of grams per mole (g/mol) and is calculated by adding the atomic masses of the elements that make up the compound as found on the periodic table.

Calculating Molar Mass

For instance, to find the molar mass of \( \mathrm{Mg(OH)_2} \), you would sum the atomic masses of Magnesium (Mg), Oxygen (O), and Hydrogen (H), taking into account the number of atoms of each in the compound.
Understanding molar mass allows students to determine how many moles of a substance are present in a given mass, and vice versa. This is crucial for titration calculations because it links mass (a measurable quantity) with moles (a chemical quantity used in stoichiometry), as demonstrated in the provided solutions.
As reflected in the textbook problem, knowing the molar mass of compounds involved can help you perform the necessary conversions to find the amounts required to reach the endpoint of a titration.