Question

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\text { aq) requires } 31.15 \mathrm{mL}\) of \(0.2420 \mathrm{M}\) KOH for titration to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

Short answer

The molarity of the H₃PO₄ solution is calculated by first finding the moles of KOH, applying the 1:2 stoichiometric relationship with H₃PO₄ at the second equivalence point, and then using the molarity formula with the obtained moles of H₃PO₄ and the provided volume of the H₃PO₄ solution. By doing these calculations, the molarity of the H₃PO₄ can be obtained.

Step-by-step solution

Calculate moles of KOH

Use the molarity formula M=n/V to find the moles of KOH. Here, M (molarity) is \(0.2420 \mathrm{M}\), and V (volume) is 31.15 mL or 0.03115 L. So, n (moles) = M × V = \(0.2420 \mathrm{M} \times 0.03115\) L.

Apply stoichiometric relationship

Because the stoichiometry established by the second equivalence point is 2 KOH : 1 H₃PO₄, the moles of H₃PO₄ is half the amount of KOH. So, divide the moles of KOH obtained in Step 1 by 2.

Calculate molarity of H₃PO₄

With the obtained moles of H₃PO₄ and the provided volume of the H₃PO₄ solution (25.00 mL or 0.02500 L), use the molarity formula M=n/V once again to calculate the molarity of the H₃PO₄ solution.

Key concepts

Titration

Titration is an analytical technique used to determine the concentration of an unknown solution. It involves the gradual addition of a known concentration solution, called the titrant, to the unknown solution until the reaction reaches the endpoint. The endpoint is typically indicated by a color change due to an indicator or through an instrumentation method such as pH meter.

During titration, the reaction between the titrant and the substance being analyzed must be known, as well as stoichiometrically exact so that measurements can be precise. For example, in the provided exercise, KOH is added to a solution of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) until the second equivalence point, which refers to the complete reaction of the second of the three acidic protons of \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantities and proportions of reactants and products in chemical reactions. It is based on the conservation of mass, where the total mass of the reactants equals the total mass of the products in a chemical reaction.

To solve a stoichiometric problem, one must understand the mole concept, balanced chemical equations, and the relationship between the reacting quantities of each substance. In the context of the exercise, the balanced equation shows that two moles of KOH react with one mole of \(\mathrm{H}_{3} \mathrm{PO}_{4}\), which is crucial for accurately determining the molarity of the phosphoric acid solution.

Equivalence Point

The equivalence point in a titration is the point at which the quantity of titrant added is exactly enough for stoichiometric reaction with the analyte; it's the point where moles of titrant equal the moles of substance of interest in the solution. This is particularly important for polyprotic acids like \(\mathrm{H}_{3} \mathrm{PO}_{4}\), which can have multiple equivalence points corresponding to the successive loss of its acidic protons.

In the given problem, reaching the second equivalence point implies that the second proton of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) has completely reacted. Identifying the equivalence point is often aided by an appropriate indicator or sophisticated equipment like a potentiometer.

Molar Concentration

Molar concentration, commonly known as molarity, is the number of moles of a solute per liter of solution. It is expressed in the unit moles per liter (M). To calculate molarity, you can use the formula \( M = \frac{n}{V} \), where \( n \) is the number of moles of the solute and \( V \) is the volume of the solution in liters.

Molarity is a critical concept for the exercise, as the final step to determine the unknown concentration of the \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution. Afterwards, the stoichiometric relationship established by the second equivalence point allows us to find the moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and hence calculate its molarity. Here, an understanding of molarity is even more important due to the dilution of the titrant (KOH) and the presence of multiple protons in \(\mathrm{H}_{3} \mathrm{PO}_{4}\).