Problem 12 - Chemical Principles: The Quest for Insight

Question

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?

Step-by-Step Solution

Answer

The molarity of H3PO4 in the solution is approximately 0.386 M and the mass of H3PO4 in the initial solution is approximately 0.385 grams.

Step 1: Calculate the Molarity after Dilution

Use the dilution formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume. Given M1 = 3.0 M and V1 = 10.0 mL, find M2 using V2 = 250 mL.

Step 2: Determine Moles of KOH Used

The molarity from Step 1 will be used along with the volume of KOH used in the titration (38.5 mL) to calculate the moles of KOH that reacted, using the formula moles = Molarity x Volume (in liters).

Step 3: Calculate Moles of H3PO4

From the balanced chemical equation, the molar ratio between KOH and H3PO4 is 3:1. Use this ratio to calculate the moles of H3PO4 that reacted.

Step 4: Calculate H3PO4 Molarity

Use the moles of H3PO4 obtained in Step 3 and the initial volume of the phosphoric acid solution (10.0 mL) to find the molarity by dividing moles by volume (in liters).

Step 5: Calculate Mass of H3PO4

Using the moles for H3PO4 from Step 3, multiply by its molar mass (98.00 g/mol) to find the mass of H3PO4.

Key Concepts

Molarity and Dilution

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 of a Chemical Reaction

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.

Mole Concept in Chemistry

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.

Calculating Molar Mass

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.