In an acid-base titration, \(25.22 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) were used to neutralize \(35.22 \mathrm{~mL}\) of \(0.313 \mathrm{M} \mathrm{NaOH}\). What was the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution? $$ \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O} $$

Understanding a balanced chemical equation is crucial in any acid-base titration problem. A balanced equation shows the relationship between the reactants and products in a chemical reaction. For our example, the balanced chemical equation is \[\mathrm{H}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O} \] which tells us that one mole of \[\mathrm{H}_{2} \mathrm{SO}_{4} \] reacts with two moles of \[\mathrm{NaOH} \]. Always ensure that the number of atoms of each element is the same on both sides of the equation. This is what makes it 'balanced'. In our case, we have:

• 1 sulfur (S) atom on both sides
• 2 sodium (Na) atoms on both sides
• 4 oxygen (O) atoms on both sides
• 4 hydrogen (H) atoms on both sides

This balance is key to accurately converting moles of one substance to moles of another using stoichiometry.

The mole ratio is derived from the coefficients in the balanced chemical equation. It provides a proportionate relationship between the amounts of reactants and products. In our balanced equation, the mole ratio is: \[\mathrm{H}_{2} \mathrm{SO}_{4} : \mathrm{NaOH} = 1:2 \] This means one mole of \[\mathrm{H}_{2} \mathrm{SO}_{4} \] reacts with two moles of \[\mathrm{NaOH} \]. This ratio helps us determine how much of one reactant is needed to completely react with another. For example, if we have 0.01102 moles of \[\mathrm{NaOH} \], we divide by 2 to find moles of \[\mathrm{H}_{2} \mathrm{SO}_{4} \]: \[\text{Moles of } \mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.01102}{2} = 0.00551 \text{ moles} \] This step simplifies calculations and ensures the correct proportions of reactants.

Molarity (M) is a measure of how many moles of a solute are in one liter of solution. The molarity formula is: \[\text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}} \] Given that we have 0.00551 moles of \[\mathrm{H}_{2} \mathrm{SO}_{4} \] and a volume of 0.02522 liters (remember to convert mL to liters by dividing by 1000): \[\text{Molarity of } \mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.00551}{0.02522} = 0.218 \mathrm{M} \] So, the molarity of the \[\mathrm{H}_{2} \mathrm{SO}_{4} \] solution is 0.218 M. This step requires accurate measurement and conversion of volumes, as well as correct usage of the molarity formula.

Neutralization reactions occur between an acid and a base to form a salt and water. In our example, \[\mathrm{H}_{2} \mathrm{SO}_{4} \], a strong acid, reacts with \[\mathrm{NaOH} \], a strong base: \[\mathrm{H}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O} \] The hydrogen ions \[\left( \mathrm{H}^{+} \right) \] from the acid react with the hydroxide ions \[\left( \mathrm{OH}^{-} \right) \] from the base to form water. The remaining ions combine to form the salt \[\mathrm{Na}_{2} \mathrm{SO}_{4} \]. Neutralization is key in titrations because it helps determine the unknown concentration (molarity) of a solution by observing the volume of the titrant used to reach the endpoint, where neutralization is complete.