A 10.0-mL volume of \(3.0 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\) is transferred to a \(250-\mathrm{mL}\) volumetric flask and diluted to the mark. It was found that \(38.5 \mathrm{~mL}\) of this diluted solution was needed to reach the stoichiometric point in a titration of \(10.0 \mathrm{~mL}\) of a phosphoric acid solution according to the reaction $$ \begin{aligned} 3 \mathrm{KOH}(\mathrm{aq})+& \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow \\ & \mathrm{K}_{3} \mathrm{PO}_{4}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(1) \end{aligned} $$ (a) Calculate the molarity of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in the solution. (b) What mass of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is in the initial solution?

Molarity is one of the most fundamental concepts in chemistry, defined as the number of moles of a substance per liter of solution. Knowing molarity is crucial for the preparation and use of solutions in laboratory practices, including titrations.

When it comes to dilution, the process doesn't change the number of moles of solute—just the total volume of the solution. The key equation used in dilution calculations is the dilution formula, represented as: \[ M_1V_1 = M_2V_2 \], where \(M_1\) and \(M_2\) are the molarities before and after dilution, and \(V_1\) and \(V_2\) are the respective volumes. Dilution calculations allow you to find the molarity of a diluted solution, which is exactly what needs to be done before progressing with titration calculations.

Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products involved in a chemical reaction. It relies on the balanced chemical equation, which provides the molar ratio of the substances involved.

In an acid-base titration, stoichiometry helps us understand the reaction proportions. For the given reaction with KOH and \(H_3PO_4\), the balanced equation tells us that it takes 3 moles of KOH to neutralize 1 mole of \(H_3PO_4\). Using the concept of stoichiometry and the molarity of solutions involved, you can calculate the amounts of acids and bases at the stoichiometric (equivalence) point during a titration.

The mole concept is essential for quantifying matter in chemical reactions. One mole represents 6.022×10^23 particles (Avogadro’s number) of a substance. In other words, it is a bridge between the atomic world and the macroscopic world we can measure.

In a titration, you often need to find out how many moles of a reactant you have to know how much of another reactant it can completely react with. For instance, by finding the number of moles of KOH using its molarity and volume in liters, we can calculate the moles of \(H_3PO_4\) needed for the reaction. Understanding the mole concept allows us to transition from volumes of titrant used to actual amounts of substance involved.

Molar mass is the mass of one mole of a substance and it has units of grams per mole (g/mol). The molar mass of a compound can be found by summing the molar masses of its constituent elements, each multiplied by the number of times that element appears in the chemical formula.

Knowing the molar mass of \(H_3PO_4\), which is 98.00 g/mol, we can calculate the mass of this substance by multiplying the molar mass with the calculated moles from the titration data. This is a critical step, as it gives the mass of the acid in the original solution, connecting quantitative data from titration to practical applications, such as determining concentration in a sample.