Question

A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires \(47.21 \mathrm{~mL}\) of a hydrochloric acid solution to react completely. $$ \mathrm{HCl}(a q)+\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ Calculate the molarity of the hydrochloric acid solution.

Short answer

The molarity of the hydrochloric acid solution is \(0.02560 \, M\).

Step-by-step solution

Convert mass of sodium hydrogen carbonate to moles

Given the mass of sodium hydrogen carbonate (NaHCO3) as 0.1015g, we can convert it into moles using its molar mass. The molar mass of NaHCO3 can be calculated as: Na (22.99 g/mol) + H (1.01 g/mol) + C (12.01 g/mol) + 3 * O (3 * 16.00 g/mol) = 84.007 g/mol Now, we can convert the mass of NaHCO3 into moles: moles of NaHCO3 = mass / molar mass = 0.1015 g / 84.007 g/mol = 1.2093×10⁻³ mol

Determine moles of hydrochloric acid

From the balanced equation, we know that the reaction between NaHCO3 and HCl is a 1:1 stoichiometry: HCl(aq) + NaHCO3(s) -> NaCl(aq) + H2O(l) + CO2(g) Thus, the moles of HCl that react are equal to the moles of NaHCO3: moles of HCl = moles of NaHCO3 = 1.2093×10⁻³ mol

Calculate the molarity of hydrochloric acid

Now that we have the moles of HCl, we can determine the molarity (M) using the volume of hydrochloric acid given. The volume of HCl is given as 47.21 mL (converting to L by dividing by 1000): Volume of HCl = 47.21 mL = 0.04721 L Molarity (M) is defined as the moles of solute per liter of solution: Molarity (M) = moles of HCl / volume of HCl = 1.2093×10⁻³ mol / 0.04721 L = 0.02560 M Thus, the molarity of the hydrochloric acid solution is 0.02560 M.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Simply put, stoichiometry is like a recipe for chemistry, providing the ratios of how much of each substance is needed to make the reaction work.

In our exercise, stoichiometry helps us understand the relationship between sodium hydrogen carbonate (NaHCO3) and hydrochloric acid (HCl). The stoichiometry of the reaction is 1:1, meaning one mole of NaHCO3 reacts with one mole of HCl. By knowing this ratio, we can confidently say that the amount of HCl used is equal to the amount of NaHCO3 reacted, allowing us to calculate the molarity of the HCl solution.

Chemical Reaction

A chemical reaction involves transforming one or more substances into new substances through the breaking and forming of chemical bonds. In this specific case, the reaction between sodium hydrogen carbonate and hydrochloric acid is a type of acid-base reaction.

The equation \(\mathrm{HCl}(aq) + \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g)\) illustrates the reactants transforming into salt, water, and carbon dioxide. Understanding the process of this chemical reaction is essential for interpreting the stoichiometric relationships and is foundational for solving the problem related to molarity.

Molar Mass

Molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It's a critical factor in stoichiometry as it allows us to convert mass to moles, which is necessary to apply the stoichiometric ratios of a balanced chemical equation.

In the exercise, the first step requires calculating the molar mass of sodium hydrogen carbonate by adding the molar masses of its constituent elements. With the molar mass known, we can then convert the given mass of NaHCO3 to moles, which is a stepping stone to finding the molarity of the HCl solution.