Question

The volume of an adult's stomach ranges from about \(50 \mathrm{~mL}\) when empty to \(1 \mathrm{~L}\) when full. If the stomach volume is \(400 \mathrm{~mL}\) and its contents have a pH of 2 , how many moles of \(\mathrm{H}^{+}\)does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\)comes from \(\mathrm{HCl}\), how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

Short answer

To totally neutralize the stomach acid in a stomach with a volume of \(400 \mathrm{~mL}\) and a pH of \(2\), \(0.336 \mathrm{~g}\) of sodium hydrogen carbonate (\(\mathrm{NaHCO_3}\)) is required.

Step-by-step solution

Find the molarity of \(\mathrm{H}^{+}\) ions

To determine the molarity of \(\mathrm{H}^{+}\) ions in the stomach, we can use the pH formula, where \(\mathrm{pH}=-\log[\mathrm{H}^+]\). Given that the pH is \(2\), we can solve for the molarity of \(\mathrm{H}^{+}\) ions: \[2 = -\log[\mathrm{H}^+]\] Now, we will find the concentration of \(\mathrm{H}^{+}\) ions by taking the antilog: \[[\mathrm{H}^+] = 10^{-2}\ \mathrm{M}\] The molarity of \(\mathrm{H}^{+}\) ions is \(10^{-2}\) M.

Calculate the number of moles and mass of \(\mathrm{NaHCO_3}\) required

First, we need to find the number of moles of \(\mathrm{H}^{+}\) ions in the stomach with a volume of \(400 \mathrm{~mL}\). We can do this by multiplying the volume in liters by the molarity: Number of moles of \(\mathrm{H}^{+}\) ions = Volume (in liters) × Molarity Number of moles of \(\mathrm{H}^{+}\) ions = \(0.4 \mathrm{~L} \times 10^{-2} \mathrm{M}\) Number of moles of \(\mathrm{H}^{+}\) ions = \(4 \times 10^{-3} \) moles Since the \(\mathrm{H}^{+}\) ions come from \(\mathrm{HCl}\), we know that the number of moles of \(\mathrm{H}^{+}\) ions is equal to the number of moles of \(\mathrm{HCl}\). Now, let's write the balanced chemical equation for the neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaHCO_3}\): \[\mathrm{HCl} + \mathrm{NaHCO_3} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2}\] From the balanced equation, we see that \(1\) mole of \(\mathrm{HCl}\) reacts with \(1\) mole of \(\mathrm{NaHCO_3}\). Thus, the number of moles of \(\mathrm{NaHCO_3}\) required for neutralization equals the number of moles of \(\mathrm{H}^{+}\) ions, which is \(4 \times 10^{-3} \) moles. Now we can calculate the mass of \(\mathrm{NaHCO_3}\) required by multiplying the number of moles by the molar mass of \(\mathrm{NaHCO_3}\): Mass of \(\mathrm{NaHCO_3}\) = Number of moles × Molar mass Molar mass of \(\mathrm{NaHCO_3} = 22.99 (\mathrm{Na}) + 1.01 (\mathrm{H}) + 12.01 (\mathrm{C}) + 16.00 \times 3 (\mathrm{O}) = 84.01\ \mathrm{g/mol}\) Mass of \(\mathrm{NaHCO_3}\) = \(4 \times 10^{-3}\) moles × \(84.01\ \mathrm{g/mol}\) Mass of \(\mathrm{NaHCO_3}\) = \(0.336\ \mathrm{g}\) Therefore, to totally neutralize the stomach acid, \(0.336\ \mathrm{g}\) of sodium hydrogen carbonate is required.

Key concepts

pH Calculation

Understanding the concept of pH is crucial for comprehending how acidity is quantified within solutions such as stomach acid. The pH scale ranges from 0 to 14, where a pH of 7 is neutral, values less than 7 are acidic, and values greater than 7 are basic or alkaline. The pH is a logarithmic measure, which means that each whole pH value below 7 is ten times more acidic than the next higher value.

For stomach acid, a pH of 2 indicates a highly acidic environment. Using the formula \(\mathrm{pH}=-\log[\mathrm{H}^+]\), we can calculate the concentration of hydrogen ions, \(\mathrm{H}^+\), in the stomach. With a pH of 2, the molarity of \(\mathrm{H}^+\) ions is \(10^{-2}\) molar (M), which gives a clear quantitive measure of the acid's strength. This molarity is the starting point for determining the amount of neutralizing agent required to balance the acidity, a common task in chemistry related to stoichiometry.

Molarity of H+ Ions

Molarity is a term used in chemistry to describe the concentration of a substance in a solution. It's defined as the number of moles of a solute divided by the volume of the solution in liters. In the context of stomach acid, molarity refers to the concentration of \(\mathrm{H}^+\) ions. When the exercise states that the stomach contents have a pH of 2, using the provided formula and calculation, we find that the concentration of \(\mathrm{H}^+\) ions is \(10^{-2}\) M.

For a solution with a volume of 400 mL, or 0.4 L, the number of moles of \(\mathrm{H}^+\) can be calculated by multiplying the volume by the molarity (\(0.4 \mathrm{~L} \times 10^{-2} \mathrm{M} = 4 \times 10^{-3} \) moles). This step is a critical part of solving stoichiometry problems in acid-base reactions, providing the fundamental information to determine the amount of reactant needed for neutralization.

Stoichiometry of Neutralization

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. In neutralization reactions, such as when stomach acid (\(\mathrm{HCl}\)) is neutralized by sodium hydrogen carbonate (\(\mathrm{NaHCO_3}\)), stoichiometry allows us to predict the amounts of substances required to reach neutrality.

The stoichiometric coefficients in the balanced chemical equation \(\mathrm{HCl} + \mathrm{NaHCO_3} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2}\) inform us that one mole of \(\mathrm{HCl}\) reacts with one mole of \(\mathrm{NaHCO_3}\). Knowing the number of moles of \(\mathrm{H}^+\) ions, we can determine the equivalent amount of \(\mathrm{NaHCO_3}\) needed to achieve a neutral solution. With stoichiometry, we translate the mole ratio from the balanced equation into a real-world quantity, allowing precise adjustments to a highly acidic environment, such as the stomach, by calculating the exact mass of \(\mathrm{NaHCO_3}\) required for complete neutralization.