91Ó°ÊÓ

In chemistry, an acid-base reaction involves an acid reacting with a base to produce water and a salt. This process is essential in many titration techniques where the goal is to determine the concentration of an unknown substance. For our problem involving the acid \( \text{H}_2\text{A} \), the acid is fully titrated with \( \text{NaOH} \), a strong base. This reaction follows the balanced equation:

• \( \text{H}_2\text{A(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{A(aq)} + 2\text{H}_2\text{O(}\ell\text{)} \)

Here, \( \text{NaOH} \) reacts completely with \( \text{H}_2\text{A} \) to yield sodium salt \( \text{Na}_2\text{A} \) and water. Understanding this chemical behavior is crucial when using titration to determine unknown concentrations. The stoichiometry or mole ratio in this reaction is 1:2 (acid to base), meaning two moles of \( \text{NaOH} \) are required for every mole of \( \text{H}_2\text{A} \).
This fundamental understanding of acid-base reactions allows us to accurately perform titrations and calculate important properties such as molar mass.

Calculating molar mass in titration can be straightforward with the right data. The molar mass is the mass of one mole of a substance, and for our acid \( \text{H}_2\text{A} \), this is our ultimate goal. The formula to find molar mass \( M \) is:

• \( M = \frac{\text{mass of the substance}}{\text{moles of the substance}} \)

In this exercise, we've determined the mass of the acid as \( 0.954 \text{ g} \) and calculated the moles from titration to be \( 0.00917 \text{ mol} \). By applying the formula:
\[ M = \frac{0.954 \text{ g}}{0.00917 \text{ mol}} \approx 104.06 \text{ g/mol} \]
This result shows the molar mass of the unknown acid \( \text{H}_2\text{A} \). Molar mass calculation forms the basis for identifying substances and understanding their chemical properties.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using the coefficients from the balanced equation to link the quantities of different substances. In our titration problem:

• For every 1 mole of \( \text{H}_2\text{A} \), 2 moles of \( \text{NaOH} \) are consumed.

This ratio is derived directly from the balanced chemical equation. From Step 3, we determined the relationship using:

• \( n_{\text{H}_2\text{A}} = \frac{n_{\text{NaOH}}}{2} \)

By calculating the moles of \( \text{NaOH} \) and using the stoichiometric coefficients, we accurately found the moles of \( \text{H}_2\text{A} \). This demonstrates stoichiometry's role in chemical calculations, helping us conform to the law of conservation of mass by accurately predicting product forms from given reactants.

Balancing chemical equations ensures that for any given reaction, the mass and number of atoms are conserved. It is fundamental to making accurate quantitative predictions in chemistry. In our problem, the chemical equation is already balanced:

• \( \text{H}_2\text{A(aq)} + 2 \text{NaOH(aq)} \rightarrow \text{Na}_2\text{A(aq)} + 2 \text{H}_2\text{O(}\ell\text{)} \)

The balance shows that we have 2 moles of \( \text{NaOH} \) reacting with 1 mole of \( \text{H}_2\text{A} \) to produce sodium salt and water. This means that all atoms present in the reactants are also present in the products, with none being lost or gained.
Being able to balance equations allows us to work out stoichiometric relationships and provides the groundwork needed to perform calculations like those for titrations and molar mass determination. Without balancing, none of the subsequent steps in our analysis would be accurate.