Problem 73 - Chemistry and Chemical Reactivity

Question

You have \(0.954 \mathrm{g}\) of an unknown acid, \(\mathrm{H}_{2} \mathrm{A},\) which reacts with 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 mI. of \(0.509 \mathrm{M} \mathrm{NaOH}\) is required to titrate the acid to the equivalence point, what is the molar mass of the acid?

Step-by-Step Solution

Answer

The molar mass of the acid H₂A is 104.10 g/mol.

Step 1: Calculate Moles of NaOH

First, calculate the moles of NaOH used in the reaction. Use the molarity formula, which is moles = molarity × volume (in liters). The volume needs to be converted from mL to L: \[ \text{Volume in Liters} = \frac{36.04 \, \text{mL}}{1000} = 0.03604 \, \text{L} \]Now, calculate the moles of NaOH:\[ \text{Moles of NaOH} = 0.509 \, \text{M} \times 0.03604 \, \text{L} = 0.01834 \, \text{mol} \]

Step 2: Relate Moles of NaOH to Moles of H₂A

The balanced equation shows that 2 moles of NaOH react with 1 mole of H₂A. Therefore, to find the moles of H₂A, divide the moles of NaOH by 2:\[ \text{Moles of } \mathrm{H}_{2} \mathrm{A} = \frac{0.01834 \, \text{mol NaOH}}{2} = 0.00917 \, \text{mol} \, \mathrm{H}_{2} \mathrm{A} \]

Step 3: Calculate Molar Mass of H₂A

Now that we have the moles of H₂A and the mass, we can calculate the molar mass. Use the molar mass formula, which is molar mass = mass/moles:\[ \text{Molar Mass of } \mathrm{H}_{2} \mathrm{A} = \frac{0.954 \, \text{g}}{0.00917 \, \text{mol}} = 104.10 \, \text{g/mol} \]

Key Concepts

Moles Calculation

Understanding moles calculation is crucial in titration problems. Moles represent the amount of substance and are key in determining how reactants participate in a chemical reaction. In this exercise, we start by calculating the moles of NaOH used. The molarity formula, \( \text{moles} = \text{molarity} \times \text{volume (in liters)} \), helps find the moles of NaOH. For example, converting the volume from milliliters to liters gives us \( 0.03604 \text{ L} \). Applying the formula:

• Moles of NaOH = \( 0.509 \text{ M} \times 0.03604 \text{ L} = 0.01834 \text{ mol} \).

This calculation shows the precise amount of NaOH used in the reaction. Understanding this helps you relate it to other reactants like \( \mathrm{H}_{2} \mathrm{A} \). Remember, getting the volume right is essential for accurate moles calculation.

Molar Mass Determination

Molar mass provides the mass of one mole of a substance and is important in identifying the unknown compound's identity. Once you know the moles of the substance, like \( \mathrm{H}_{2} \mathrm{A} \), and the mass you have, you can calculate its molar mass using the formula: \( \text{molar mass} = \frac{\text{mass}}{\text{moles}} \).

• For \( \mathrm{H}_{2} \mathrm{A} \), with a mass of \( 0.954 \text{ g} \) and moles \( 0.00917 \text{ mol} \), the molar mass is:
• \( \frac{0.954 \text{ g}}{0.00917 \text{ mol}} = 104.10 \text{ g/mol} \).

This calculation tells us the weight of one mole of \( \mathrm{H}_{2} \mathrm{A} \). Figuring out molar mass can help determine the molecular formula and better understand the compound you're dealing with.

Balanced Chemical Equation

A balanced chemical equation is essential in understanding how reactants convert to products. It shows the ratio of moles required for each reactant and product. In the given reaction, \( \mathrm{H}_{2} \mathrm{A}(\mathrm{aq}) + 2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{A}(\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \), for every molecule of \( \mathrm{H}_{2} \mathrm{A} \), two molecules of \( \mathrm{NaOH} \) are needed.

• This ratio allows us to calculate the moles of \( \mathrm{H}_{2} \mathrm{A} \) from the moles of \( \mathrm{NaOH} \):
• \( \text{Moles of } \mathrm{H}_{2} \mathrm{A} = \frac{0.01834 \text{ mol NaOH}}{2} = 0.00917 \text{ mol} \).

Understanding balanced equations helps you grasp how substances interact in chemical reactions. It's a fundamental skill in any chemistry problem.