Question

If the titration of a 10.0 -mL sample of sulfuric acid requires \(28.15 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium hydroxide, what is the molarity of the acid? $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Short answer

The molarity of the sulfuric acid is approximately 0.141 M.

Step-by-step solution

Write Down the Titration Reaction

The balanced chemical equation for the reaction given is \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text{aq}) + 2 \mathrm{NaOH}(\text{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\text{aq}) + 2 \mathrm{H}_{2} \mathrm{O}(\text{l})\). This equation shows that 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with 2 moles of \(\mathrm{NaOH}\).

Calculate Moles of NaOH

The volume of sodium hydroxide used in the titration is \(28.15\mathrm{~mL}\) or \(0.02815\mathrm{~L}\). Using the molarity of \(\mathrm{NaOH}\), which is \(0.100\mathrm{~M}\), calculate the moles by multiplying molarity by volume: \( n_{\mathrm{NaOH}} = 0.100 \times 0.02815 = 0.002815 \) moles of \(\mathrm{NaOH}\).

Relate Moles of NaOH to H2SO4

According to the reaction, 2 moles of \(\mathrm{NaOH}\) are needed for 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Therefore, the moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are half those of \(\mathrm{NaOH}\): \( n_{\mathrm{H}_{2} \mathrm{SO}_{4}} = \frac{0.002815}{2} = 0.0014075 \) moles.

Calculate Molarity of H2SO4

The volume of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) used is \(10.0\mathrm{~mL}\) or \(0.010\mathrm{~L}\). Molarity is calculated as moles divided by volume: \( M = \frac{0.0014075}{0.010} = 0.14075 \mathrm{~M}\).

Finalize the Solution

The calculated molarity of the sulfuric acid solution is approximately \(0.141\mathrm{~M}\).

Key concepts

Molarity Calculation

In chemistry, molarity is a convenient way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. This allows you to know how "strong" or "weak" a solution is. For instance, if you have a solution with a higher molarity, it means there are more solute particles in the solution, making it more concentrated. To calculate molarity, you use the formula: \[ M = \frac{n}{V} \] Where:

• \(M\) is the molarity,
• \(n\) is the number of moles of solute,
• \(V\) is the volume of the solution in liters.

Let's apply this to sulfuric acid in the given exercise. We determined that the number of moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(0.0014075\) moles, and the volume is \(0.010 \mathrm{~L}\). Plugging these into the formula gives a molarity \(M = \frac{0.0014075}{0.010} = 0.14075 \mathrm{~M}\). This result tells us the sulfuric acid has a concentration of around \(0.141 \mathrm{~M}\).

Stoichiometry

Stoichiometry involves using relationships between reactants and products in chemical equations to determine quantities like moles and masses. It is a crucial concept in chemistry that helps us understand how substances interact in a reaction and how much of each substance is needed or produced. In the titration problem, stoichiometry comes into play to convert the moles of sodium hydroxide \(\mathrm{NaOH}\) used into moles of sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\). According to the balanced chemical equation, \(1\) mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with \(2\) moles of \(\mathrm{NaOH}\). So, if you know the moles of \(\mathrm{NaOH}\) used were \(0.002815\) moles, the moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) needed will be half of that amount. This leads to the relationship:\[ n_{\mathrm{H}_{2} \mathrm{SO}_{4}} = \frac{n_{\mathrm{NaOH}}}{2} = \frac{0.002815}{2} = 0.0014075 \text{ moles}\] By understanding the proportional relationships in the chemical equation, stoichiometry simplifies the calculations needed in chemical analysis.

Chemical Equation Balancing

Chemical equation balancing is a fundamental aspect of chemistry that ensures the same number of each kind of atom is present among all reactants and products. It's like ensuring both sides of the scale are equal. The balanced chemical equation for our titration is: \[ \mathrm{H}_{2} \mathrm{SO}_{4}(aq) + 2 \mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(aq) + 2 \mathrm{H}_{2} \mathrm{O}(l) \] This equation shows that one molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two molecules of \(\mathrm{NaOH}\) to form one molecule of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and two molecules of water. Balancing this equation involved ensuring the atoms for each element are equal on both sides. Let's break it down:

• The equation starts with 2 hydrogen and 1 sulfate ions in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and matches it with 2 hydroxide ions from 2 \(\mathrm{NaOH}\).
• The sodium ions combine with the sulfate ion to form \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), while the leftover hydrogen reacts to form \(2\) water molecules.

This balance ensures the law of conservation of mass is met, critical for accurate chemical reactions, as otherwise, the calculations for reactants and products would be incorrect.