Problem 118 - Chemistry The Central Science In Si Units

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 \(\mathrm{pH}\) of 2 , how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

Step-by-Step Solution

Answer

0.336 grams of sodium hydrogen carbonate is needed.

Step 1: Calculate Moles of Hydrogen Ions

The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration, given by the formula: \[ \text{pH} = -\log_{10}[\mathrm{H}^+] \] Given that the pH is 2, we can find the concentration of hydrogen ions, \[ [\mathrm{H}^+] = 10^{-\text{pH}} = 10^{-2} \text{ M} \] The stomach volume is 400 mL (or 0.4 L), so the number of moles of \( \mathrm{H}^+ \) is: \[ \text{Moles of } \mathrm{H}^+ = 0.4 \text{ L} \times 10^{-2} \text{ M} = 4 \times 10^{-3} \text{ moles} \]

Step 2: Calculate Moles of Sodium Hydrogen Carbonate Required

To neutralize the acid, sodium hydrogen carbonate \( \mathrm{NaHCO}_3 \) reacts with hydrochloric acid \( \mathrm{HCl} \) as follows: \[ \mathrm{HCl} + \mathrm{NaHCO}_3 \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \] This equation indicates a 1:1 mole ratio for \( \mathrm{HCl} \) and \( \mathrm{NaHCO}_3 \). Therefore, \[ 4 \times 10^{-3} \text{ moles of } \mathrm{NaHCO}_3 \] are needed to neutralize \( 4 \times 10^{-3} \text{ moles of } \mathrm{H}^+ \).

Step 3: Find Grams of Sodium Hydrogen Carbonate

The molar mass of sodium hydrogen carbonate \( \mathrm{NaHCO}_3 \) is approximately 84 g/mol. Using this molar mass, we calculate the mass of \( \mathrm{NaHCO}_3 \) needed: \[ \text{Mass of NaHCO}_3 = 4 \times 10^{-3} \text{ moles} \times 84 \text{ g/mol} = 0.336 \text{ g} \]

Step 4: Final Answer

Therefore, 0.336 grams of sodium hydrogen carbonate is required to neutralize the stomach acid when the stomach volume is 400 mL with a pH of 2.

Key Concepts

pH Calculations

Understanding pH calculations is crucial in chemistry, especially when dealing with acid-base reactions.
The pH of a solution is a measure of its acidity or basicity. It is calculated using the formula:

• \[pH = -\log_{10}[\mathrm{H}^+]\]

A low pH value indicates a high concentration of hydrogen ions, meaning the solution is acidic. Conversely, a high pH means low hydrogen ion concentration, indicating a basic solution.
In our exercise, we have a solution with a pH of 2, which signifies a highly acidic environment. The calculation involves converting this pH value into the hydrogen ion concentration by taking the inverse log base 10 of the negative pH. This gives us the formula:

• \[[\mathrm{H}^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M}\]

This concentration tells us how many hydrogen ions are present per liter of solution.

Hydrogen Ion Concentration

The hydrogen ion concentration is a pivotal part of understanding the nature of an acidic solution. In any solution, the concentration of hydrogen ions can reveal its acidity level.
This is particularly important in the context of physiological conditions like those in a human stomach. The hydrogen ion concentration can be directly derived from the pH value as demonstrated:

• \[[\mathrm{H}^+] = 10^{-\text{pH}}\]

For our specific scenario with the stomach content at pH 2:

• The hydrogen ion concentration is \[10^{-2} \, \text{M}\].

Given the volume of the stomach content as 400 mL (or 0.4 L), you can calculate the total moles of hydrogen ions by multiplying this concentration with the volume:

• \[\text{Moles of } \mathrm{H}^+ = 0.4 \, \text{L} \times 10^{-2} \, \text{M} = 4 \times 10^{-3} \, \text{moles}\]

This calculation provides insight into the amount of acid present in a quantified manner.

Stoichiometry Calculations

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It plays a significant role in determining how much of a substance is required or produced in a reaction.
In the problem at hand, we use stoichiometry to calculate the amount of sodium hydrogen carbonate (baking soda) necessary to neutralize the stomach acid. The reaction equation is:

• \[\mathrm{HCl} + \mathrm{NaHCO}_3 \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\]

This balanced equation shows that one mole of hydrochloric acid (\( \mathrm{HCl} \)) reacts with one mole of sodium bicarbonate (\( \mathrm{NaHCO}_3 \)). Therefore, the stoichiometric calculation shows:

• For \( 4 \times 10^{-3} \) moles of \( \mathrm{H}^+ \), you need \( 4 \times 10^{-3} \) moles of \( \mathrm{NaHCO}_3 \).

Knowing the molar mass of \( \mathrm{NaHCO}_3 \) is about 84 g/mol, the mass required is given by:

• \[\text{Mass of NaHCO}_3 = 4 \times 10^{-3} \text{ moles} \times 84 \text{ g/mol} = 0.336 \text{ g}\]

Thus, stoichiometry helps determine that 0.336 grams of sodium hydrogen carbonate are needed to neutralize the stomach acid fully.