Question

In an acid-base titration, 45.78 \(\mathrm{mL}\) of a sulfuric acid solution is titrated to the end point by 74.30 \(\mathrm{mL}\) of 0.4388 \(\mathrm{M}\) sodium hydroxide solution. What is the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Short answer

The molarity of the H\(_2\)SO\(_4\) solution is 0.3559 M.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is: H\(_2\)SO\(_4\) + 2 NaOH → Na\(_2\)SO\(_4\) + 2 H\(_2\)O This tells us that one H\(_2\)SO\(_4\) molecule reacts with 2 NaOH molecules. Therefore, the mole ratio between H\(_2\)SO\(_4\) and NaOH is 1:2.

Calculate moles of NaOH

We have the volume (74.30 mL) and molarity (0.4388 M) of sodium hydroxide solution. We can calculate the moles of NaOH using the formula: Moles = Volume (in liters) × Molarity First, convert the volume from mL to L: 74.30 mL × (1 L / 1000 mL) = 0.0743 L Now, calculate moles of NaOH: Moles of NaOH = 0.0743 L × 0.4388 M = 0.03258 moles of NaOH

Calculate moles of H\(_2\)SO\(_4\)

Using the mole ratio we found in Step 1 (1:2), we can calculate the moles of H\(_2\)SO\(_4\) reacting with the moles of NaOH: Moles of H\(_2\)SO\(_4\) = (0.03258 moles of NaOH) × (1 mole of H\(_2\)SO\(_4\) / 2 moles of NaOH) = 0.01629 moles of H\(_2\)SO\(_4\)

Find the molarity of H\(_2\)SO\(_4\) solution

Now that we have the moles of H\(_2\)SO\(_4\), we can find its molarity. We know that the volume of H\(_2\)SO\(_4\) solution is 45.78 mL. First, convert the volume from mL to L: 45.78 mL × (1 L / 1000 mL) = 0.04578 L Now, we can find the molarity of the H\(_2\)SO\(_4\) solution using the formula: Molarity = Moles / Volume (in liters) Molarity of H\(_2\)SO\(_4\) = 0.01629 moles / 0.04578 L = 0.3559 M The molarity of the H\(_2\)SO\(_4\) solution is 0.3559 M.

Key concepts

Chemical Stoichiometry

Chemical stoichiometry involves the calculation of the relative quantities of reactants and products in chemical reactions. It is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products in a chemical reaction.

Understanding stoichiometry is crucial when dealing with reactions like the neutralization between an acid and a base. In the context of our acid-base titration calculation, the balanced equation \[\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}\]is the starting point. It shows that sulfuric acid (\(\text{H}_2\text{SO}_4\)) reacts with sodium hydroxide (\(\text{NaOH}\)) in a 1:2 mole ratio. This information is essential to finding the relationship between the moles of each reactant and product, allowing us to calculate the unknown concentration.

Molarity

Molarity is a measurement of concentration, representing the number of moles of a solute that is present in one liter of solution. It is denoted as \(\text{M}\) and is calculated using the formula:\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\].

In the acid-base titration example, molarity enables us to find the moles of sodium hydroxide based on its concentration and volume. It also allows us to compute the concentration of sulfuric acid after determining the moles of acid that reacted. Since titration relies on accurately known concentrations, molarity plays a pivotal role in identifying the endpoint of the reaction where the proportions of acid and base align perfectly with the stoichiometric ratios.

Mole Ratio

The mole ratio is derived from the coefficients of a balanced chemical equation, and it provides the proportional relationship between reactants and products. In the case of the given titration problem, the mole ratio is gleaned from the balanced equation:\[\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}\].

This equation establishes that for every mole of sulfuric acid, two moles of sodium hydroxide are required for a complete reaction. The mole ratio (1:2) is integral in linking the moles of sodium hydroxide used in the titration to the moles of sulfuric acid. This connection lets us calculate the desired molarity of the sulfuric acid solution by incorporating the moles of acid and the volume of the acid solution.