Question

A solution of \(0.162\) M \(\mathrm{NaOH}\) is used to neutralize \(25.0 \mathrm{~mL}\) of a \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. If \(32.8 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution is required, what is the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution? \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The molarity of the H2SO4 solution is 0.106 M.

Step-by-step solution

- Write the balanced chemical equation

Note the balanced chemical equation: $$ \mathrm{H}_2 \mathrm{SO}_4(aq) + 2\, \mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_2 \mathrm{SO}_4(aq) + 2\, \mathrm{H}_2 \mathrm{O}(l) $$.

- Calculate the moles of NaOH

Use the volume and molarity of the NaOH solution to find the moles of NaOH used: $$ M(\mathrm{NaOH}) = 0.162 \text{ M}, \quad V(\mathrm{NaOH}) = 32.8 \; \text{mL} \quad ( \text{convert to liters} \ : \ 32.8 \; \text{mL} = 0.0328 \; L) $$ Calculate the moles by multiplying the molarity by the volume: $$ n(\mathrm{NaOH}) = M(\mathrm{NaOH}) \times V(\mathrm{NaOH}) = 0.162 \; \text{mol/L} \times 0.0328 \; \text{L} $$ $$ n(\mathrm{NaOH}) = 0.0053136 \; \text{mol} $$.

- Find the moles of H2SO4 neutralized

From the balanced equation, 1 mole of H2SO4 reacts with 2 moles of NaOH. Use stoichiometry to determine the moles of H2SO4: $$ n(\mathrm{H}_2 \mathrm{SO}_4) = \frac{n(\mathrm{NaOH})}{2} = \frac{0.0053136 \; \text{mol}}{2} $$ $$ n(\mathrm{H}_2 \mathrm{SO}_4) = 0.0026568 \; \text{mol} $$.

- Calculate the molarity of the H2SO4 solution

Using the volume and moles of H2SO4, calculate its molarity: $$ V(\mathrm{H}_2 \mathrm{SO}_4) = 25.0 \; \text{mL} \quad ( \text{convert to liters} \ : \ 25.0 \; \text{mL} = 0.025 \; \text{L} ) $$ $$ M(\mathrm{H}_2 \mathrm{SO}_4) = \frac{n(\mathrm{H}_2 \mathrm{SO}_4)}{V(\mathrm{H}_2 \mathrm{SO}_4)} = \frac{0.0026568 \; \text{mol}}{0.025 \; \text{L}} $$ $$ M(\mathrm{H}_2 \mathrm{SO}_4) = 0.106272 \; \text{M} $$.

Key concepts

Molarity Calculation

Molarity measures the concentration of a solution, represented as moles of solute per liter of solution (M = mol/L). To find the molarity, you need the number of moles and the volume of the solution in liters. Start by converting volumes from milliliters to liters. For example, 32.8 mL converts to 0.0328 L.
The formula to calculate moles is: \[n = M \times V\] where: \( n\) is the number of moles, \( M\) is the molarity, and \( V\) is the volume in liters. Multiply the given molarity of NaOH (0.162 M) by the converted volume (0.0328 L) to obtain the moles of NaOH (0.0053136 mol). This process allows you to determine how much of the substance is present in the solution, which is crucial for further steps like stoichiometry.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It relies on balanced chemical equations to show the proportionate relationships of reactants and products. In the neutralization reaction, the balanced equation: \[\mathrm{H}_2\mathrm{SO}_4 (aq) + 2 \; \mathrm{NaOH} (aq) \rightarrow \; \mathrm{Na}_2\mathrm{SO}_4(aq) + 2 \; \mathrm{H}_2\mathrm{O} (l)\] shows that 1 mole of \(\mathrm{H}_2\mathrm{SO}_4\) reacts with 2 moles of \(\mathrm{NaOH}\).
Using stoichiometry, you can relate the moles of \(\mathrm{NaOH}\) to the moles of \(\mathrm{H}_2\mathrm{SO}_4\). Half the moles of \(\mathrm{NaOH}\) (0.0053136 mol) yield the moles of \(\mathrm{H}_2\mathrm{SO}_4\) (0.0026568 mol), because of the 1:2 mole ratio from the balanced equation. This proportional relationship simplifies calculations throughout multiple steps in chemistry problems.

Chemical Equations

Chemical equations represent what happens in a chemical reaction, showing reactants turning into products. Equations must be balanced, with the same number of each type of atom on both sides, following the Law of Conservation of Mass. In this neutralization reaction: \[\mathrm{H}_2\mathrm{SO}_4 (aq) + 2 \; \mathrm{NaOH} (aq) \rightarrow \; \mathrm{Na}_2\mathrm{SO}_4 (aq) + 2 \; \mathrm{H}_2\mathrm{O} (l)\] each element has equal quantities on both sides. Balancing chemical equations involves adjusting coefficients to match the atom counts, ensuring the reaction keeps mass constant. This fundamental ensures accurate stoichiometry and reliable chemical calculations.

Acid-Base Reaction

Acid-base reactions involve the transfer of protons (H\(^+\)) between reactants. Acids donate protons; bases accept them. In the example, \(\mathrm{H}_2\mathrm{SO}_4\) (sulfuric acid) reacts with \(\mathrm{NaOH}\) (sodium hydroxide). The balanced equation is: \[\mathrm{H}_2\mathrm{SO}_4(aq) + 2 \; \mathrm{NaOH}(aq) \rightarrow \; \mathrm{Na}_2\mathrm{SO}_4(aq) + 2 \; \mathrm{H}_2 \mathrm{O}(l)\] The sulfuric acid donates two protons to the sodium hydroxide, which accepts them, producing water and sodium sulfate. This neutralization reaction leads to water and salt (\(\mathrm{Na}_2\mathrm{SO}_4\)), demonstrating how acids and bases neutralize each other. Such reactions are key in many chemical processes, from titrations to practical applications in everyday life.