Question

An antacid tablet contains sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), and inert ingredients. A 0.465-g sample of powdered tablet was mixed with \(53.3 \mathrm{~mL}\) of \(0.190 \mathrm{M}\) \(\mathrm{HCl}\) (hydrochloric acid). The mixture was allowed to stand until it reacted. $$ \begin{aligned} \mathrm{NaHCO}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow & \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \end{aligned} $$ The excess hydrochloric acid was titrated with \(54.6 \mathrm{~mL}\) of \(0.128 \mathrm{M} \mathrm{NaOH}\) (sodium hydroxide). $$ \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What is the percentage of sodium hydrogen carbonate in the antacid?

Short answer

The percentage of sodium hydrogen carbonate in the tablet is approximately 56.66%.

Step-by-step solution

Calculate Moles of Initial HCl

Determine the initial moles of HCl added to the reaction. Use the formula: \( ext{moles} = ext{concentration (M)} \times ext{volume (L)} \). For HCl, the initial concentration is \(0.190 \text{ M}\) and the volume is \(53.3 \text{ mL} = 0.0533 \text{ L}\). Thus,\[\text{moles of } \mathrm{HCl} = 0.190 \text{ M} \times 0.0533 \text{ L} = 0.010127 \text{ moles}\text{ } \mathrm{HCl}\].

Calculate Moles of Excess HCl

Find the moles of HCl that were in excess and reacted with NaOH. Again, use \( ext{moles} = ext{concentration} \times ext{volume} \). The concentration of NaOH is \(0.128 \text{ M}\) and the volume used in the titration is \(54.6 \text{ mL} = 0.0546 \text{ L}\). Thus,\[\text{moles of } \mathrm{HCl}_{ ext{excess}} = 0.128 \text{ M} \times 0.0546 \text{ L} = 0.0069888 \text{ moles}\text{ } \mathrm{HCl}\].

Calculate Moles of HCl Reacted with NaHCO3

Subtract the moles of excess HCl from the initial moles to find the amount that reacted with \(\mathrm{NaHCO}_{3}\): \[\text{moles of } \mathrm{HCl}_{\text{reacted with NaHCO}_3} = 0.010127 - 0.0069888 = 0.0031382 \text{ moles}\text{ } \mathrm{HCl}\].

Calculate Moles of NaHCO3 in Tablet

The balanced equation shows a 1:1 molar ratio between \(\mathrm{NaHCO}_{3}\) and \(\mathrm{HCl}\). Thus, the moles of \(\mathrm{NaHCO}_{3}\) are the same as the moles of HCl reacted:\[\text{moles of } \mathrm{NaHCO}_{3} = 0.0031382 \text{ moles}\text{ } \mathrm{NaHCO}_{3}\].

Calculate Mass of NaHCO3 in Tablet

Use the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \) to convert moles to grams. The molar mass of \(\mathrm{NaHCO}_{3}\) is approximately \(84.01 \text{ g/mol}\). Thus,\[\text{mass of } \mathrm{NaHCO}_{3} = 0.0031382 \text{ moles} \times 84.01 \text{ g/mol} = 0.2635 \text{ g}\].

Calculate Percentage of NaHCO3 in Tablet

Divide the mass of \(\mathrm{NaHCO}_{3}\) by the total mass of the tablet, then multiply by 100 to find the percentage:\[\text{Percentage of } \mathrm{NaHCO}_{3} = \left( \frac{0.2635}{0.465} \right) \times 100 \approx 56.66\%\].

Key concepts

Stoichiometry

In chemistry, stoichiometry relates to the quantitative relationships in chemical reactions. It allows us to predict the amounts of reactants needed and the amounts of products formed. In the context of the solution provided, stoichiometry was essential in understanding the relationship between the reactants, sodium hydrogen carbonate (\(\mathrm{NaHCO}_{3}\)) and hydrochloric acid (\(\mathrm{HCl}\)). Each reactant's moles were calculated to analyze their consumption and to find out which reactant was limiting. The balanced chemical equation reveals a 1:1 molar ratio. This means each mole of \(\mathrm{NaHCO}_{3}\) reacts with exactly one mole of \(\mathrm{HCl}\). Hence, the stoichiometric coefficients directly influenced the calculated moles of reactants and dictated the procedural flow in the steps.

Acid-Base Reaction

Acid-base reactions are fundamental in chemistry and involve the transfer of protons between reactants. In this exercise, the key reaction is between \(\mathrm{NaHCO}_{3}\), a base, and \(\mathrm{HCl}\), an acid. When these two react, they form sodium chloride (\(\mathrm{NaCl}\)), water (\(\mathrm{H}_{2}\mathrm{O}\)), and carbon dioxide (\(\mathrm{CO}_{2}\)).

• This type of reaction demonstrates how acids can neutralize bases, forming water and a salt.
• The removal of \(\mathrm{CO}_{2}\) as a gas further drives the reaction to completion.
• Understanding the concept of conjugate acids and bases is essential, as \(\mathrm{HCl}\) is a strong acid, fully dissociating to donate protons.

In addition, the titration step further involved an acid-base reaction between the excess \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) to determine the amount of reacted and excess \(\mathrm{HCl}\). This adds a practical application to the theoretical understanding of acid-base reactions.

Molarity

Molarity is a way of expressing the concentration of a solute in a solution, defined as the moles of solute per liter of solution. In this exercise, molarity was crucial for determining the initial amount of \(\mathrm{HCl}\) and the excess \(\mathrm{HCl}\) that reacted with \(\mathrm{NaOH}\). By knowing the molarity and the volume of the solutions used, one could find the exact moles involved in reactions:

• Initial moles of \(\mathrm{HCl}\): 0.190 M and 53.3 mL (0.0533 L) lead to 0.010127 moles
• Excess \(\mathrm{HCl}\): Calculated via \(\mathrm{NaOH}\) titration with 0.128 M and 54.6 mL (0.0546 L) resulting in 0.0069888 moles.

Molarity provides a direct way to segue from volume to mole calculations, reinforcing the foundation for quantitative chemical analysis.

Mole Calculations

Moles are a basic unit used to quantify chemical substances. They connect macroscopic amounts to the atomic scale. In this exercise, mole calculations were continuously used to track substances and understand reaction dynamics.

• At the start, we converted molarity and volume of \(\mathrm{HCl}\) to initial moles.
• We calculated how much \(\mathrm{HCl}\) reacted and how much was excess.
• Through stoichiometry, these values directly told us the moles of \(\mathrm{NaHCO}_{3}\) involved.

Once determined, we converted moles back into mass for \(\mathrm{NaHCO}_{3}\) using its molar mass. Moving seamlessly between these chemical quantities underpins practical problem-solving in chemistry and bridges between theoretical concepts and real-world applications.