Problem 155 - General Chemistry

Question

The active ingredients of an antacid tablet contained only magnesium hydroxide and aluminum hydroxide. Complete neutralization of a sample of the active ingredients required \(48.5 \mathrm{~mL}\) of \(0.187 M\) hydrochloric acid. The chloride salts from this neutralization were obtained by evaporation of the filtrate from the titration; they weighed \(0.4200 \mathrm{~g}\). What was the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet?

Step-by-Step Solution

Answer

The mass percentage of magnesium hydroxide is 27.77%.

Step 1: Understand the Reaction

The neutralization reaction involves magnesium hydroxide \((Mg(OH)_2)\) and aluminum hydroxide \((Al(OH)_3)\) reacting with hydrochloric acid \((HCl)\). The balanced reactions are: \[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \] \[ Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O \] These reactions help determine stoichiometry between reactants and products.

Step 2: Calculate Moles of HCl

The volume and molarity of hydrochloric acid can be used to find its moles. \[ C = \frac{n}{V} \Rightarrow n = C \times V \]\[ n_{HCl} = 0.187 \text{ M} \times 0.0485 \text{ L} = 0.0090695 \text{ moles} \] This is the total moles of HCl used for complete reaction.

Step 3: Find Moles of Resulting Chlorides

From the reactions, moles of HCl can be accounted for by the amount of resulting chloride salts. From stoichiometry:- For each mole of \(Mg(OH)_2\), 2 moles of \(HCl\) are used, producing 1 mole of \(MgCl_2\).- For each mole of \(Al(OH)_3\), 3 moles of \(HCl\) are used, producing 1 mole of \(AlCl_3\).

Step 4: Set Up Equations

Let \(x\) be the moles of \(Mg(OH)_2\) and \(y\) the moles of \(Al(OH)_3\). Based on stoichiometry:\[ 2x + 3y = 0.0090695 \quad \text{(Equation for HCl)} \]

Step 5: Consider Mass of Chloride Salts

Using molar masses, calculate the total mass of the resulting chloride salts:- Molar mass of \(MgCl_2 = 95.21 \text{ g/mol}\)- Molar mass of \(AlCl_3 = 133.33 \text{ g/mol}\)The mass from the salts:\[ 95.21x + 133.33y = 0.4200 \quad \text{grams} \]

Step 6: Solve Simultaneous Equations

Solve the two equations:1. \(2x + 3y = 0.0090695\)2. \(95.21x + 133.33y = 0.4200\)By solving, \(x = 0.002\) and \(y = 0.0017\).

Step 7: Calculate Mass of Mg(OH)2

Find the mass of \(Mg(OH)_2\) using the number of moles and its molar mass (58.32 g/mol):\[ \text{Mass of } Mg(OH)_2 = 0.002 \times 58.32 = 0.11664 \text{ g} \]

Step 8: Calculate Percentage of Mg(OH)2

Calculate the percentage by mass of \(Mg(OH)_2\):\[ \text{Percentage} = \left(\frac{\text{Mass of } Mg(OH)_2}{\text{Total Mass of Active Ingredients}}\right) \times 100 = \left(\frac{0.11664}{0.4200}\right) \times 100 = 27.77\% \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, involving calculations based on balanced chemical equations. It allows us to understand the quantitative relationships between reactants and products in a chemical reaction.
In acid-base titrations, stoichiometry helps us determine how much reactant is needed to completely react with another. For example, in our given reactions, we have:\[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \] and \[ Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O \].
From these reactions, stoichiometry provides the ratio of how many moles of hydrochloric acid (HCl) are needed to completely react with magnesium hydroxide (Mg(OH)_2) and aluminum hydroxide (Al(OH)_3). Specifically:

• For Mg(OH)_2: 1 mole of Mg(OH)_2 requires 2 moles of HCl.
• For Al(OH)_3: 1 mole of Al(OH)_3 requires 3 moles of HCl.

Understanding stoichiometry is crucial for solving the simultaneous equations later in the problem.

Molarity

Molarity is an important concept in acid-base titration, as it describes the concentration of a solution. It is denoted by the symbol 'M' and is defined as the number of moles of solute per liter of solution. The formula to calculate molarity (C) is:
\[ C = \frac{n}{V} \] where \(n\) is the number of moles, and \(V\) is the volume in liters.
In the given problem, the hydrochloric acid solution has a molarity of 0.187 M. Given that 48.5 mL (or 0.0485 L) was used, the moles of HCl can be found using: \[ n_{HCl} = 0.187 \times 0.0485 = 0.0090695 \text{ moles} \] This represents the total amount of HCl used to achieve complete neutralization of the antacid components.

Neutralization Reaction

Neutralization reactions are chemical reactions where an acid and a base react to form water and a salt. In the context of the problem, the antacid ingredients, magnesium hydroxide (Mg(OH)_2) and aluminum hydroxide (Al(OH)_3), act as bases. They react with hydrochloric acid (HCl), an acid, to neutralize each other.
This process occurs as per the reactions:

• \( Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \)
• \( Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O \)

Neutralization reactions are used in titration to determine the concentration of an unknown solution. They are also crucial in understanding chemical reactions involved in everyday medications like antacids, which aim to neutralize stomach acid effectively.

Molar Mass

Molar mass is another critical concept in solving the problem. It represents the mass of one mole of a chemical substance, usually expressed in grams per mole (g/mol). To find the mass of reaction components or products, we use their molar masses.
In this problem, after the neutralization reaction, chloride salts formed include magnesium chloride (MgCl_2) and aluminum chloride (AlCl_3). Their molar masses are:

• \( MgCl_2 = 95.21 \text{ g/mol} \)
• \( AlCl_3 = 133.33 \text{ g/mol} \)

Using molar mass, we relate moles of each chloride to its mass to connect stoichiometric calculations with physical mass measurements. For instance, solving for moles of \( Mg(OH)_2 \) and \( Al(OH)_3 \), and multiplying them by their respective molar masses provides the mass contribution of each chloride in total yield, further aiding the percentage composition calculation for magnesium hydroxide in antacids.