Question

A $6.50-g$ sample of a diprotic acid requires $137.5\mathrm{mL}$ of a $0.750M$ NaOH solution for complete neutralization. Determine the molar mass of the acid.

Short answer

The molar mass of the diprotic acid is approximately $126\text{ g/mol}$.

Step-by-step solution

Determine the amount of moles of NaOH used for the reaction

We need to find the amount of moles of NaOH involved in the reaction by multiplying the volume of the solution by its molar concentration. The volume must be converted to liters first. Volume of NaOH in L = $137.5\text{ mL}$ × $\frac{1\text{ L}}{1000\text{ mL}}$ = $0.1375\text{ L}$ Amount of moles of NaOH = volume × concentration = $0.1375\text{ L}$ × $0.750\text{ M}$ = $0.1031\text{ mol}$

Calculate the amount of moles of the diprotic acid

Since the ratio of the diprotic acid to NaOH in the reaction is 1:2 (due to the acid being diprotic and donating 2 protons), we can divide the amount of moles of NaOH by 2 to find the amount of moles of the diprotic acid. Amount of moles of diprotic acid = $\frac{0.1031\text{ mol NaOH}}{2}$ = $0.0516\text{ mol}$

Calculate the molar mass of the diprotic acid

Now that we have the amount of moles of the diprotic acid and its mass, we can find its molar mass by dividing the mass by the amount of moles. Molar mass of diprotic acid = $\frac{6.50\text{ g}}{0.0516\text{ mol}}$ = $126\text{ g/mol}$ So, the molar mass of the diprotic acid is approximately $126\text{ g/mol}$.

Key concepts

Mole Concept

Understanding the mole concept is essential for diving into chemistry problems, especially when dealing with chemical quantities. A mole corresponds to Avogadro's number, which is roughly 6.022×10^23 particles, atoms, or molecules of a substance. It's a standard unit in chemistry that provides a bridge between the atomic world and the macroscopic world we observe.

For instance, when we refer to a 6.50-g sample of a diprotic acid in our exercise, we’re thinking about how many molecules of that acid we have. To find this out, we convert that mass to moles, which then allows us to understand the amount in relation to a substance's molecular or formula weight. In simpler terms, the mole concept is a counting tool used by chemists to relate masses of substances to the actual number of particles.

Acid-Base Neutralization

Acid-base neutralization is the reaction where an acid and a base react quantitatively with each other. In a typical acid-base reaction, hydrogen ions (H+) from the acid combine with hydroxide ions (OH−) from the base to form water (H2O). An acid is termed 'diprotic' if it can donate two protons per molecule during the reaction, which is why we needed twice the amount of NaOH to react completely with the diprotic acid in our exercise.

In neutralization reactions, it's vital to determine the stoichiometry, which will help us understand the proportions in which the reactants combine. For a diprotic acid, the stoichiometry tells us that it takes two moles of NaOH to neutralize one mole of the acid.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the substances as they participate in various chemical reactions. It enables us to predict the products and reactants involved in a chemical reaction based on the law of conservation of mass.

In the exercise, stoichiometry allows us to deduce that since the diprotic acid gives away two protons (H+ ions), it will require two moles of NaOH to neutralize one mole of the acid, leading us to comprehend that the relationship between the moles of the acid and NaOH is 1:2. By applying these ratios, stoichiometry helps us calculate the necessary amount of each reactant needed for a reaction to go to completion, as seen in the step-by-step solution where we found that 0.1031 moles of NaOH relate to only 0.0516 moles of the acid.

Molarity

Molarity is a measure of the concentration of a solution and is defined as the number of moles of solute per liter of solution. It's often represented by the symbol 'M' and is of crucial importance in preparing solutions for reactions and in our case, understanding the interaction between an acid and a base in a neutralization reaction.

For instance, when we talk about a '0.750 M NaOH solution', we are saying that for every liter of this solution, there are 0.750 moles of NaOH. Molarity can be used to find the number of moles present in any volume of solution by multiplying the molarity by the volume in liters, as perfectly demonstrated in the solved example where we multiplied the volume of NaOH (0.1375 L) by 0.750 M to find the moles of NaOH.