Problem 97 - Chemistry The Molecular Science

Question

You are given \(0.954 \mathrm{~g}\) of an unknown acid, \(\mathrm{H}_{2} \mathrm{~A},\) which reacts with \(\mathrm{NaOH}\) according to the balanced equation $$\mathrm{H}_{2} \mathrm{~A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}_{2} \mathrm{~A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ If \(36.04 \mathrm{~mL}\) of \(0.509-\mathrm{M} \mathrm{NaOH}\) is required to react with all of the acid, calculate the molar mass of the acid.

Step-by-Step Solution

Answer

The molar mass of the acid is approximately 104.08 g/mol.

Step 1: Calculate moles of NaOH

First, determine the number of moles of \( \mathrm{NaOH} \). Use the formula for moles: \[ \text{moles of NaOH} = \text{concentration} \times \text{volume} \] Here, the volume in liters is \( 36.04 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.03604 \text{ L} \). Thus, \[ \text{moles of NaOH} = 0.509 \text{ M} \times 0.03604 \text{ L} = 0.01833636 \text{ moles} \]

Step 2: Determine moles of H2A

According to the balanced equation, 2 moles of \( \mathrm{NaOH} \) react with 1 mole of \( \mathrm{H}_{2} \mathrm{~A} \). Thus, the moles of \( \mathrm{H}_{2} \mathrm{~A} \) is half the moles of \( \mathrm{NaOH} \): \[ \text{moles of } \mathrm{H}_{2} \mathrm{~A} = \frac{0.01833636 \text{ moles NaOH}}{2} = 0.00916818 \text{ moles} \]

Step 3: Calculate the molar mass of H2A

Molar mass is the mass of a substance divided by the number of moles. Given the mass of \( \mathrm{H}_{2} \mathrm{~A} \) is \( 0.954 \text{ g} \), use the formula: \[ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.954 \text{ g}}{0.00916818 \text{ moles}} \approx 104.08 \text{ g/mol} \]

Key Concepts

Molar Mass Calculation

Calculating the molar mass is an essential skill in stoichiometry and is vital to understanding reactions quantitatively. Molar mass is the mass of one mole of a given substance, expressed in grams per mole (g/mol). To find it, divide the sample's mass by the number of moles it contains.
In our exercise, the unknown acid \( \mathrm{H}_2 \mathrm{A} \) was given to us with a mass of \( 0.954 \text{ g} \). Before calculating the molar mass, we needed to determine how many moles of \( \mathrm{H}_2 \mathrm{A} \) were present. From our balanced chemical reaction, we found \( 0.00916818 \text{ moles} \) of the acid. Finally, the molar mass formula, \[ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} \] gave us \( 104.08 \text{ g/mol} \) as the molar mass of the acid. Mastering this calculation lets you convert between mass and moles easily, providing a deeper insight into chemical processes. Keep practicing, and it will soon become second nature!

Acid-Base Titration

Acid-base titration is a method used to determine the concentration of an unknown acid or base solution. In the process, a neutralization reaction occurs between an acid and a base to form water and a salt. This technique is fundamental in analytical chemistry, offering precise quantitative analysis.
In our example, titration helped identify how much \( \mathrm{NaOH} \) was needed to completely react with the given unknown acid, \( \mathrm{H}_2 \mathrm{A} \). We had \( 36.04 \text{ mL} \) of a \( 0.509 \text{ M} \) sodium hydroxide solution. Using the volume and concentration, we calculated the moles of \( \mathrm{NaOH} \) as \( 0.01833636 \text{ moles} \).

• The goal of the titration was to determine the point at which an equivalent amount of acid and base had reacted, known as the equivalence point.
• At this stage, the number of moles of \( \mathrm{NaOH} \) added is stoichiometrically equivalent to half the amount of \( \mathrm{H}_2 \mathrm{A} \).

Mastering titration ensures you can effectively determine unknown concentration values in various chemical processes. It's a classic technique you'll encounter frequently, so understanding it thoroughly is beneficial!

Balanced Chemical Equation

A balanced chemical equation is a fundamental concept that represents a chemical reaction with an equal number of each type of atom on both sides of the equation. This balance reflects the law of conservation of mass, ensuring that matter is neither created nor destroyed during the reaction process.
In our specific exercise, the balanced equation was: \[ \mathrm{H}_2 \mathrm{~A} (\mathrm{aq}) + 2 \mathrm{NaOH} (\mathrm{aq}) \rightarrow \mathrm{Na}_2 \mathrm{~A} (\mathrm{aq}) + 2 \mathrm{H}_2 \mathrm{O} (\ell) \].

• Each component in this reaction is crucially balanced so that the number and type of atoms remain constant throughout the process.
• The equation reflects the stoichiometric relationship between reactants and products, highlighting that two moles of \( \mathrm{NaOH} \) are required to react with one mole of \( \mathrm{H}_2 \mathrm{~A} \)

Understanding how to accurately balance chemical equations is essential. It not only provides insights into the proportions of substances involved in a reaction but also into the reaction's potential yield. Balancing equations serves as the foundation for any stoichiometric calculation, ensuring precision and accuracy in chemical quantification.