Problem 71 - 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 $$ \mathrm{H}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ If 36.04 mL of 0.509 M \(\mathrm{NaOH}\) is required to titrate the acid to the second equivalence point, what is the molar mass of the acid?

Step-by-Step Solution

Answer

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

Step 1: Calculate Moles of NaOH

First, we need to calculate the number of moles of NaOH using the concentration and volume. The formula to use is: \( ext{moles} = ext{concentration (M)} \times ext{volume (L)} \). Here, the concentration of NaOH is 0.509 M and the volume is 36.04 mL (which is 0.03604 L). Thus, the moles of NaOH are \( 0.509 \times 0.03604 = 0.01833636 \) moles.

Step 2: Relate Moles of NaOH to Moles of Acid

According to the balanced chemical equation, 2 moles of NaOH react with 1 mole of \( ext{H}_2 ext{A} \). Therefore, the moles of \( ext{H}_2 ext{A} \) will be half the moles of NaOH. Therefore, \( ext{moles of } ext{H}_2 ext{A} = 0.01833636 / 2 = 0.00916818 \) moles.

Step 3: Calculate Molar Mass of the Acid

The molar mass of the acid can be found using the formula: \( ext{Molar Mass} = \frac{ ext{mass of acid in grams}}{ ext{moles of acid}} \). The mass of the acid is 0.954 g, and the moles of acid are 0.00916818. Thus, \( ext{Molar Mass} = \frac{0.954}{0.00916818} = 104.08 \text{ g/mol} \).

Key Concepts

Molar Mass

Molar mass is a fundamental concept in chemistry. It represents the mass of a given substance divided by the amount of substance, typically expressed in grams per mole. To determine the molar mass of a compound, you need to have both the total mass of the compound and the number of moles involved. In the exercise, we determined the molar mass of the acid \( \mathrm{H}_{2} \mathrm{A} \) by dividing the mass of the acid, 0.954 grams, by the number of moles, 0.00916818 moles. This calculation resulted in a molar mass of 104.08 g/mol.

Knowing the molar mass is crucial for a variety of chemical calculations, such as stoichiometry and reaction yield predictions. Typically, you might be asked to find the molar mass so that you can ascertain how much of a given chemical is present in a sample or what proportion it takes in a reaction.

Balanced Equation

A balanced chemical equation is vital for any stoichiometric calculation or understanding of a chemical reaction. It shows the exact ratio in which reactants combine to form products. According to the exercise, the given reaction equation is:

\[ \mathrm{H}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \]

This equation is balanced because it has the same number of each type of atom on both sides. There are two sodium (Na) atoms, two hydroxide ions (OH), two hydrogen atoms from the acid, and a single atom of \( \mathrm{A} \) on each side of the reaction. A balanced equation ensures chemical conservation and allows us to calculate the molar relationships between substances accurately. Following this equation enables us to relate the moles of NaOH to moles of \( \mathrm{H}_{2} \mathrm{A} \) as used in stoichiometry.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. Understanding these relationships allows chemists to predict how much product will form from a given amount of reactant or determine the amount of reactant required to produce a certain amount of product.

In our exercise, stoichiometry helped relate the moles of NaOH to the moles of the acid \( \mathrm{H}_{2} \mathrm{A} \).

• The balanced equation showed that 2 moles of NaOH react with 1 mole of \( \mathrm{H}_{2} \mathrm{A} \).
• This allowed us to calculate that if \( 0.01833636 \) moles of NaOH were used, then only half that amount, or \( 0.00916818 \) moles of \( \mathrm{H}_{2} \mathrm{A} \), reacted.

Such stoichiometric calculations often employ molar ratios derived from the balanced equation, providing a bridge between substances to meaningfully convert from one to another within the reaction's context.