Problem 45 - Chemistry: Principles and Reactions

Question

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Step-by-Step Solution

Answer

Answer: The percentage of sodium hydrogen carbonate in the sample is approximately \(71.61 \%\).

Step 1: Calculate moles of \(\mathrm{H}^{+}\) ions

To calculate the moles of \(\mathrm{H}^{+}\) ions, we will use the formula: Moles of \(\mathrm{H}^{+}\) = volume (in liters) × concentration (in moles per liter) Moles of \(\mathrm{H}^{+}\) = \(\frac{15.5 \: \mathrm{mL}}{1000}\) × \(0.275 \: \mathrm{M}\) = \(0.0042625 \: \mathrm{mol}\)

Step 2: Calculate moles of \(\mathrm{NaHCO}_{3}\)

The balanced chemical equation given in the exercise states that one mole of \(\mathrm{H}^{+}\) reacts with one mole of \(\mathrm{NaHCO}_{3}\). Therefore, the moles of \(\mathrm{NaHCO}_{3}\) in the sample is equal to the moles of \(\mathrm{H}^{+}\) ions. Moles of \(\mathrm{NaHCO}_{3}\) = \(0.0042625 \: \mathrm{mol}\)

Step 3: Calculate mass of \(\mathrm{NaHCO}_{3}\) in the sample

To calculate the mass of \(\mathrm{NaHCO}_{3}\), we will use the formula: Mass of \(\mathrm{NaHCO}_{3}\) = moles of \(\mathrm{NaHCO}_{3}\) × molar mass of \(\mathrm{NaHCO}_{3}\) Molar mass of \(\mathrm{NaHCO}_{3}\) = \(23 + 1 + 12 + 16 \times 3 = 84 g/mol\) Mass of \(\mathrm{NaHCO}_{3}\) = \(0.0042625 \: \mathrm{mol} \times 84 \: \mathrm{g/mol} = 0.35805 \: \mathrm{g}\)

Step 4: Calculate the percentage of \(\mathrm{NaHCO}_{3}\) in the sample

The final step is to calculate the percentage of \(\mathrm{NaHCO}_{3}\) in the \(0.500 \: \mathrm{g}\) sample. To do this, we will use the formula: Percentage of \(\mathrm{NaHCO}_{3}\) = \(\frac{\text{Mass of } \mathrm{NaHCO}_{3}}{\text{Total mass of sample}} \times 100 \%\) Percentage of \(\mathrm{NaHCO}_{3}\) = \(\frac{0.35805 \: \mathrm{g}}{0.500 \: \mathrm{g}} \times 100 \% \approx 71.61 \%\) The percentage of sodium hydrogen carbonate in the sample is approximately \(71.61 \%\).

Key Concepts

Sodium Hydrogen Carbonate

Sodium hydrogen carbonate, also known as baking soda, is a white crystalline compound with a slightly alkaline taste. It has the chemical formula \( \mathrm{NaHCO}_{3} \). This compound is commonly used in various applications such as baking, as a leavening agent, and in medicine, where it acts as an antacid to relieve indigestion and heartburn.

When used in titration, sodium hydrogen carbonate reacts with acids, releasing carbon dioxide gas as a byproduct. The balanced reaction shows that one mole of \( \mathrm{NaHCO}_{3} \) reacts with one mole of \( \mathrm{H}^{+} \) ions to form sodium ions, carbon dioxide, and water.

• NaHCO₃ is helpful for determining the concentration of acids in a titration
• It is essential in chemical analysis due to its predictable reaction pattern

Knowing how \( \mathrm{NaHCO}_{3} \) interacts with acids makes it a valuable compound for chemistry students to study.

Stoichiometry

Stoichiometry is a core concept in chemistry that deals with the quantitative relationships between the reactants and products in chemical reactions. In the context of the given exercise, stoichiometry helps determine how much sodium hydrogen carbonate reacts with hydrochloric acid based on their molar ratios from the balanced equation.

The steps to use stoichiometry often include:

• Start by writing a balanced chemical equation
• Calculate moles of any one substance (reactant or product)
• Use the mole ratio from the balanced equation to find moles of the needed substance
• Finally, convert moles to grams if required

Here, the balanced equation provides the direct 1:1 ratio between \( \mathrm{NaHCO}_{3} \) and \( \mathrm{H}^{+} \). Understanding stoichiometry in titrations helps predict how much of each reactant is needed for the reaction to proceed completely.

Acid-Base Reaction

An acid-base reaction is a fundamental chemical reaction where an acid donates protons (\( \mathrm{H}^{+} \)) to a base, leading to the formation of water and a salt. Such reactions are essential for neutralizing solutions.

In the exercise, an acid-base reaction occurs between sodium hydrogen carbonate and hydrochloric acid where \( \mathrm{H}^{+} \) from hydrochloric acid interacts with \( \mathrm{NaHCO}_{3} \). This results in the formation of \( \mathrm{Na}^{+} \), \( \mathrm{CO}_{2} \) gas, and water.

• The equation shows that acid-base reactions often produce gases like \( \mathrm{CO}_{2} \) as a byproduct
• These reactions help to calculate the concentration of unknown solutions in titrations

Sodium hydrogen carbonate, acting as a base, accepts \( \mathrm{H}^{+} \) ions in this scenario, making it a classic example of an acid-base reaction. Understanding such interactions is key in various applications, from cooking to chemical manufacturing.