Question

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3},\) in a powder for stomach upsets is found by titrating with \(0.275 \mathrm{M}\) 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 $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Short answer

Answer: The percentage of sodium hydrogen carbonate, NaHCO₃, in the sample is approximately 71.62%.

Step-by-step solution

Calculate the moles of H⁺ ions in the titration

We are given the volume (15.5 mL) and the concentration (0.275 M) of the HCl solution used in the titration. We can use these values to find the moles of H⁺ ions. Since 1 M of HCl has 1 mole of H⁺ ions, we can use the given concentration directly to obtain moles. Moles of H⁺ ions = volume (in L) × concentration (in mol/L) Moles of H⁺ ions = (15.5 mL × (1 L/1000 mL)) × 0.275 mol/L = 0.0042625 mol

Determine the moles of NaHCO₃ in the sample

Based on the stoichiometry of the balanced equation, 1 mol of NaHCO₃ reacts with 1 mol of H⁺ ions. Therefore, the moles of NaHCO₃ in the sample are equal to the moles of H⁺ ions used in the titration. Moles of NaHCO₃ = 0.0042625 mol

Convert moles of NaHCO₃ to grams

Now, we need to convert the moles of NaHCO₃ in the sample to grams. To do this, we can use the molar mass of NaHCO₃, which is: Molar mass of NaHCO₃ = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 48.00 g/mol (3 O) = 84.01 g/mol Now, multiply the moles of NaHCO₃ by its molar mass to obtain grams: Grams of NaHCO₃ = moles of NaHCO₃ × molar mass of NaHCO₃ Grams of NaHCO₃ = 0.0042625 mol × 84.01 g/mol = 0.358080625 g

Calculate the percentage of NaHCO₃ in the sample

Finally, we can find the percentage of NaHCO₃ in the 0.500 g sample. To do this, divide the grams of NaHCO₃ by the total grams of the sample and multiply by 100: Percentage of NaHCO₃ = (grams of NaHCO₃ / total grams of sample) × 100 Percentage of NaHCO₃ = (0.358080625 g / 0.500 g) × 100 = 71.62 % The percentage of sodium hydrogen carbonate, NaHCO₃, in the sample is approximately 71.62%.

Key concepts

Titration

Titration is a precise analytical method used to determine the concentration of a known reactant in a solution. By gradually adding a titrant (in this case, hydrochloric acid, HCl) to a sample until the reaction is complete, the point of stoichiometric equivalence, or the endpoint, is reached. The volume of titrant used provides valuable information when it corresponds to a known chemical reaction equation.

During titration, chemists often use a balanced chemical equation to understand the stoichiometry of the reaction. In our exercise, the balanced equation represents the neutralization between sodium hydrogen carbonate (NaHCO₃) and H⁺ ions from hydrochloric acid. The process involves careful measurement, and through calculations based on the endpoint volume, the analyst can determine the substance's purity or concentration in a sample.

Stoichiometry

Stoichiometry is fundamental in chemistry, bridging the gap between the molecular world and measurable quantities. It involves using the coefficients of a balanced chemical reaction to determine the relative amounts of reactants and products. The stoichiometric coefficients indicate the proportion of moles that react with each other. In the provided exercise, the stoichiometry tells us that one mole of NaHCO₃ reacts with one mole of H⁺ ions.

Understanding stoichiometry is critical to performing accurate titration analyses. The molar ratio derived from the balanced equation allows us to equate moles of one substance to moles of another. This relation is essential for calculating the percentage composition of a compound in a mixture, as demonstrated in the exercise.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a vital concept when converting between moles and grams, both of which are units chemists use to measure substances. The molar mass is calculated by summing the atomic masses of all atoms in a formula unit of a compound. For instance, the molar mass of NaHCO₃ is found by adding the atomic masses of sodium (Na), hydrogen (H), carbon (C), and three oxygen atoms (O).

The accurate determination of molar mass is necessary for stoichiometry calculations in titrations. It enables chemists to convert the moles of a compound, obtained from the stoichiometry of the reaction, into a mass that can be understood and applied to practical situations, such as calculating the percentage composition of a substance in a sample.

Chemical Reaction Equation

A chemical reaction equation provides the blueprint of a chemical reaction, showing the reactants converting into products. It outlines the process in symbols and formulas, wherein each substance is represented by its chemical formula, and the coefficients indicate the number of moles of each substance involved in the reaction. A balanced chemical equation is necessary for stoichiometric calculations because it ensures that the law of conservation of mass is upheld.

For the percentage composition calculation in the exercise, the balanced chemical reaction equation was essential to understand how sodium hydrogen carbonate reacts with hydrochloric acid. The equation shows a 1:1 mole ratio between NaHCO₃ and H⁺, which is the foundation for the subsequent stoichiometric calculations. Without this equation, determining the exact amount of NaHCO₃ that reacts would not be possible.