How many grams of Al(OH) 3 are required to neutralize 150. mL of stomach acid with a pH of 1.5?

Hello. And this problem we are asked to solve for the mountain grams of magnesium hydroxide necessary to neutralize 30 millis of gastric acid with a P H of two note that gastric acid is mainly hydrochloric acid. So let's begin by writing a path of how we will solve this problem. So we're given the P H from that, we can find the hydro mine concentration. You can use the hydro num ion concentration to determine the concentration of hydrochloric acid. Given the concentration of hydrochloric acid and the volume, we can determine the moles of hydrochloric acid from the mold of hydrochloric acid. In the balanced reaction equation, we can determine the moles of magnesium hydroxide. And then we can determine the mass of magnesium hydroxide. We call the hydrochloric acid is a strong acid. So it will dissociate completely to form hydro mines and chlorine for every one mole of hydrochloric acid that dissociates, we form one mole of hydro mines. Since the hydrochloric acid goes completely deformed hydro mines. Concentration of hydrochloric acid will be equal to the concentration of hydro mines. We can determine the concentration of hydro lines from the P H the P H is equal to two, which is equal to the negative log of the hydro ion concentration. Hydro ion concentration then is equal to the inverse log. So one times 10 to the minus two molar which is equal to then the concentration of hydroponic acid writing a neutralization reaction between hydrochloric acid and magnesium hydroxide. They react then to form water, water comes from the hydron ions of the acid and the hydroxid ions of the base. And then a salt, our salt then is formed from the cat iron from our base and the an iron from our acid. So we form magnesium chloride. We then balance the reaction. We have two hydroxide ions. We need two hydros to react to the two hydroxides and we'll form two waters. We have 30 mL of gastric acid which we'll consume or assume to be hydrochloric acid. We'll convert our volume from the leaders to leaders to multiply by one leader and divide by 1000 mL our volume of hydrochloric acid in the numerator and denominator cancels will then make use of the concentration to go from volume to moles to multiply by one times 10 to minus two moles of hydrochloric acid divide by a liter. Our units of liters of hydrochloric acid and the numerator and denominator cancels it then can make use of the balanced reaction equation. We have one mole of magnesium hydroxide. So multiplied by one mole of magnesium hydroxide for every two moles of hydroponic acid. So we'll divide by two moles of hydrochloric acid. Our moles of hydrochloric acid in the numerator and denominator cancels. And then we'll convert our moles of magnesium hydroxide to mass of magnesium hydroxide using its smaller mass. So we'll multiply it by 58.320 g. And then by, by one mole of magnesium hydroxide, our moles of magnesium hydroxide in the numerator and denominator cancels. This then is equal to 8.7 times 10 to minus three g of magnesium hydroxide. This is then the mass of magnesium hydroxide. And we need to react with 30 mL of gastric acid, which we assume to be hydrochloric acid with a P H of two. Thanks for watching. Hope it's helped.

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10. Acids and Bases

Strong Acid Strong Base Titrations (Simplified)

GOB Chemistry

10. Acids and Bases

Strong Acid Strong Base Titrations (Simplified)

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3:59 minutes

Problem 70

Textbook Question

How many grams of Al(OH)₃ are required to neutralize 150. mL of stomach acid with a pH of 1.5?

Verified step by step guidance

Step 1: Calculate the concentration of hydrogen ions ([H⁺]) in the stomach acid using the pH formula: \( \text{pH} = -\log[H^+] \). Rearrange the formula to find \( [H^+] = 10^{-\text{pH}} \), and substitute \( \text{pH} = 1.5 \) to determine \( [H^+] \).

Step 2: Determine the moles of hydrogen ions in the stomach acid. Use the formula \( \text{moles of } H^+ = [H^+] \times \text{volume in liters} \). Convert the given volume of 150. mL to liters by dividing by 1000.

Step 3: Write the balanced chemical equation for the reaction between aluminum hydroxide (Al(OH)₃) and hydrogen ions (H⁺): \( \text{Al(OH)}_3 + 3H^+ \rightarrow \text{Al}^{3+} + 3H_2O \). From the equation, note that 1 mole of Al(OH)₃ reacts with 3 moles of H⁺.

Step 4: Calculate the moles of Al(OH)₃ required to neutralize the stomach acid. Use the stoichiometric ratio from the balanced equation: \( \text{moles of Al(OH)}_3 = \frac{\text{moles of } H^+}{3} \).

Step 5: Convert the moles of Al(OH)₃ to grams. Use the molar mass of Al(OH)₃ (\( \text{Al: 26.98 g/mol, O: 16.00 g/mol, H: 1.008 g/mol} \)) to calculate its molar mass, then multiply the moles of Al(OH)₃ by the molar mass to find the mass in grams.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce water and a salt. In this context, aluminum hydroxide (Al(OH)₃) acts as a base that will react with the hydrochloric acid (HCl) present in stomach acid. The balanced chemical equation for this reaction is Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O, which is essential for calculating the amount of Al(OH)₃ needed.

pH and Hydrogen Ion Concentration

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. A pH of 1.5 indicates a highly acidic solution, with a hydrogen ion concentration of approximately 0.0316 M. Understanding the relationship between pH and hydrogen ion concentration is crucial for determining how much base is needed to neutralize the acid.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced equation. It allows us to determine the exact amount of Al(OH)₃ required to neutralize the given volume of stomach acid. By using the molarity of the acid and the stoichiometric coefficients from the balanced equation, we can calculate the grams of Al(OH)₃ needed for complete neutralization.

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