Question

A 20.00 mL sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) requires \(18.67 \mathrm{mL}\) of \(0.1885 \mathrm{M} \mathrm{NaOH}\) for titration from the first 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 \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) solution is 0.0585 M.

Step-by-step solution

Interpreting the problem

It's important to understand molarity M is defined as mol/L. Also, phosphoric acid (\(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\)) has three hydrogen ions to offer for each molecule, while sodium hydroxide (\(\mathrm{NaOH}\)) can accept only one per molecule. Therefore, 1 mol of \(\mathrm{H}_{3} \mathrm{PO}_{4} (\mathrm{aq})\) will react with 3 mol of \(\mathrm{NaOH}\).

Calculating the moles of \(\mathrm{NaOH}\)

To start, calculate the number of moles of \(\mathrm{NaOH}\) added to the solution using its formula \(M = \frac{mol}{L}\). Rearranging this gives \(mol = M * L\). The molarity of the \(\mathrm{NaOH}\) solution is given as 0.1885 M and volume is given as 18.67 mL, which should be converted to liters (0.01867 L). Hence, number of moles becomes 0.1885 * 0.01867 = 0.00352 mol.

Calculating the molarity of \(\mathrm{H_{3}PO_{4}}\)

As 1 mol of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) will react with 3 mol of \(\mathrm{NaOH}\), the number of moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) will be one third of the \(\mathrm{NaOH}\) moles. Hence, the number of moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) = 0.00352 mol / 3 = 0.00117 mol. Finally, molarity can be calculated by dividing the number of moles by the volume (in liters). The volume is 20.00 mL or 0.02 L. Substituting the values, the molarity = 0.00117 mol / 0.02 L = 0.0585 M.

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that describes the concentration of a solution. It tells you how many moles of a solute are present in one liter of solution. This is expressed in units of moles per liter (mol/L). Calculating molarity helps you understand the strength of the solution and how it might react with other substances.

To find molarity, use the formula \( M = \frac{n}{V} \), where \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.

For the exercise, we determined the molarity of sodium hydroxide (NaOH) before applying this information to find the molarity of phosphoric acid (H\(_3\)PO\(_4\)). Understanding how to manipulate the molarity formula is crucial for solving such titration problems.

Phosphoric Acid

Phosphoric acid, represented chemically as \( \text{H}_3\text{PO}_4 \), is a triprotic acid. This means it can donate up to three protons (H\(^+\) ions) per molecule. This property is significant in titration because it dictates how the acid will react with bases like sodium hydroxide.

In the given problem, the reaction between phosphoric acid and sodium hydroxide involves two stages. This brings us from one equivalence point to the second. Each equivalence point signifies a stage where one mole of \( \text{H}_3\text{PO}_4 \) has reacted with sufficient \( \text{NaOH} \) to neutralize one hydrogen ion (H\(^+\)). As such, understanding how each proton reacts helps determine not just the reaction progress, but also the solution's molarity.

Sodium Hydroxide

Sodium hydroxide, or \(\text{NaOH}\), is a strong base commonly used in titrations to determine the concentration of acids. It consists of sodium ions (Na\(^+\)) and hydroxide ions (OH\(^-\)).

• \(\text{NaOH}\) is highly soluble in water and fully dissociates, making each mole of \(\text{NaOH}\) equivalent to a mole of \(\text{OH}^-\) in solution.
• When \(\text{NaOH}\) reacts with acids like phosphoric acid, it accepts hydrogen ions (H\(^+\)) from the acid, forming water (H\(_2\)O) and neutralizing the acid.

In our problem, knowing the amount of \(\text{NaOH}\) and its molarity was necessary to calculate how much was needed to reach the second equivalence point when titrating \(\text{H}_3\text{PO}_4\). This allowed us to determine the molarity of the phosphoric acid.

Equivalence Point

The equivalence point in a titration is a critical moment when the quantity of titrant added is exactly sufficient to react with the analyte in the solution.

For phosphoric acid, which has three dissociable hydrogens, there are three equivalence points, each representing a stage where one proton is fully reacted with the base. In this exercise, moving from the first to the second equivalence point indicated the progression from just one to two protons being neutralized by \(\text{NaOH}\).

At the second equivalence point, the reaction involved the complete neutralization of two hydrogen ions from \(\text{H}_3\text{PO}_4\) by \(\text{NaOH}\). Understanding equivalence points is crucial in determining the composition and required titrant volume in titration experiments.