Question

72\. Stomach acid is approximately \(0.020 \mathrm{M}\) HCl. What volume of this acid is neutralized by an antacid tablet that weighs \(330 \mathrm{mg}\) and contains \(41.0 \% \mathrm{Mg}(\mathrm{OH})_{2}, 36.2 \%\) \(\mathrm{NaHCO}_{3}\), and \(22.8 \% \mathrm{NaCl}\) ? The reactions involved are $$ \begin{aligned} \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{H}^{+}(a q) & \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

Short answer

Answer: The antacid tablet neutralizes approximately 0.303 liters (303 mL) of 0.020 M stomach acid.

Step-by-step solution

Calculate the mass of Mg(OH)2 and NaHCO3 in the tablet.

We are given the percentage of Mg(OH)2 and NaHCO3 in the antacid tablet, so we can find their mass by taking the percentage of the total mass of the tablet (330 mg). Mass of Mg(OH)2 = 330 mg * 41.0% = 330 mg * 0.41 = 135.3 mg Mass of NaHCO3 = 330 mg * 36.2% = 330 mg * 0.362 = 119.46 mg

Find the number of moles of Mg(OH)2 and NaHCO3.

Now that we have the mass of Mg(OH)2 and NaHCO3, we can find the number of moles by dividing the mass by their respective molar masses. Molar mass of Mg(OH)2 = 24.31 g/mol (Mg) + 2 * (16.00 g/mol (O) + 1.01 g/mol (H)) &= 58.33 g/mol Molar mass of NaHCO3 = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3 * 16.00 g/mol (O) = 84.01 g/mol Number of moles of Mg(OH)2 = (135.3 mg) / (58.33 g/mol) = 0.00232 mol Number of moles of NaHCO3 = (119.46 mg) / (84.01 g/mol) = 0.00142 mol

Calculate the moles of HCl required to neutralize Mg(OH)2 and NaHCO3.

We are given the chemical reactions for the neutralization process, from which we can determine the stoichiometry of the reactants. For Mg(OH)2: 1 mol of Mg(OH)2 reacts with 2 mol of HCl For NaHCO3: 1 mol of NaHCO3 reacts with 1 mol of HCl Moles of HCl required to neutralize Mg(OH)2 = 2 * moles of Mg(OH)2 = 2 * 0.00232 mol = 0.00464 mol Moles of HCl required to neutralize NaHCO3 = moles of NaHCO3 = 0.00142 mol Total moles of HCl required = moles of HCl for Mg(OH)2 + moles of HCl for NaHCO3 = 0.00464 mol + 0.00142 mol = 0.00606 mol

Calculate the volume of 0.020 M HCl neutralized by the tablet.

To find the volume of stomach acid neutralized by the tablet, we can use the given concentration of HCl (0.020 M) and the moles of HCl required. Volume = moles of HCl / concentration = 0.00606 mol / 0.020 M = 0.303 L So, the volume of stomach acid neutralized by the antacid tablet is approximately 0.303 liters (303 mL).

Key concepts

Stoichiometry

Stoichiometry plays a pivotal role in chemistry, especially when studying reactions, such as neutralization reactions. It is the study of the quantitative relationships, or ratios, between substances as they participate in chemical reactions. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction.

For instance, in a reaction between an acid and a base, stoichiometry can help determine how much base is required to neutralize a given amount of acid. This is crucial in reactions involving antacids, where the objective is to neutralize stomach acid. In the exercise provided, stoichiometry is used to calculate the volume of stomach acid that can be neutralized by the active ingredients in an antacid tablet. Once the number of moles of each reactant is known, stoichiometric coefficients from the balanced chemical equations guide us to determine the total moles of acid that can be neutralized.

Acid-Base Titration

Acid-base titration is a method used in chemistry to determine the concentration of an unknown acid or base solution by neutralizing it with a base or acid of known concentration. The point at which neutralization occurs is called the equivalence point and is often indicated by a color change with the aid of an indicator.

Determining the Endpoint in Titration

In a typical acid-base titration, a known volume of an acid or base with an unknown concentration is placed into a flask, and a titrant (a solution of known concentration) is added drop by drop until the reaction reaches the endpoint. The volume of titrant used gives information about the stoichiometry of the reaction and thus allows for the calculation of the unknown concentration.

In the provided exercise, the concept of titration is applied through the reactions of Mg(OH)₂ and NaHCO₃ with HCl. The exercise essentially asks us to perform a 'theoretical titration' by calculating how much HCl is needed to reach the equivalence point with the given amounts of antacid components.

Molar Mass

The molar mass is a fundamental concept in stoichiometry, defined as the mass of one mole of a substance. It provides a bridge between the macroscopic amount of substance you can measure—mass in grams—and the microscopic amount of substance at the molecular or atomic level—moles. Molar masses of compounds are calculated by summing the molar masses of the individual elements constituting the compound based on its chemical formula.

The molar mass is essential for converting mass to moles, which allows for the use of stoichiometry in reactions. In the example from the textbook, the molar mass of Mg(OH)₂ and NaHCO₃ is required to determine the number of moles of both substances. These moles then play a key role in calculating how much stomach acid can be neutralized by the antacid tablet.