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

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is like a recipe for chemistry, providing the proportions of each substance involved. In the context of our problem with the diprotic acid, stoichiometry allows us to relate the amount of sodium hydroxide (NaOH), which is the base used for the neutralization, to the amount of acid present in the sample.

To understand stoichiometry, it's essential to grasp the concept of the mole, which is a unit that measures the amount of substance. One mole contains Avogadro’s number of particles, which is approximately 6.022 x 10^23 particles. In stoichiometry, we use the coefficients in the balanced chemical equation to form ratios, called stoichiometric coefficients, which tell us how many moles of one substance react with another.

In our diprotic acid problem, once we establish how many moles of NaOH react, stoichiometry allows us to calculate the moles of the diprotic acid by using the mole ratio. Simply put, if two moles of NaOH are needed for one mole of acid, then we can find the moles of acid by dividing the moles of NaOH by two, as demonstrated in the steps provided.

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. It is a specific example of a double displacement reaction. The general form for the neutralization of an acid by a base is:\begin{align*}\t\t\t\t\t\t\t\t\t\t\t\t\t\t \text{Acid} + \text{Base} \rightarrow \text{Water} + \text{Salt}\text{\}\text{\}\text{\}\text{\}\text{\}\text{\}\text{In this exercise, we are looking at the neutralization of a diprotic acid (an acid that can donate two protons or hydrogen ions) by sodium hydroxide. The reaction can be represented as:}\text{\( H_2A + 2NaOH \rightarrow 2H_2O + Na_2A \)}\text{The generated salt here is sodium salt of the acid’s conjugate base (Na_2A), confirming a complete neutralization. This balanced reaction is fundamental because it shows the exact ratio in which the reactants combine, which is crucial for calculating the molar mass of the unknown acid.}

Balancing chemical equations is the process of ensuring that the number of atoms of each element is the same on both sides of the equation. This reflects the law of conservation of mass, which states that matter can neither be created nor destroyed in a chemical reaction.In our exercise, the balanced equation for the neutralization of the diprotic acid by NaOH is:\text{\( H_2A + 2NaOH \rightarrow 2H_2O + Na_2A \)}This indicates that each molecule of the diprotic acid (\text{\(H_2A\)}) reacts with two molecules of sodium hydroxide (\text{\(NaOH\)}), producing two molecules of water (\text{\(H_2O\)}) and one molecule of the sodium salt of the conjugate base (\text{\(Na_2A\)}). Ensuring the equation is balanced is vital for achieving accurate stoichiometric calculations. Without balancing, we cannot correctly interpret or predict the outcomes of the chemical reaction.

Molarity is a measure of the concentration of a solution, indicating how many moles of solute are present in one liter of solution. It's commonly used in chemistry to quantify the strength of a solution and is represented by the symbol 'M'. The formula to calculate molarity is:\text{\( \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters (L)}} \)}In our example, the molarity of the NaOH solution is given as 0.750 M, which means there are 0.750 moles of NaOH in one liter of solution. To find the moles of NaOH used for the neutralization, we multiply the volume of the solution used (in liters) by its molarity. Once we have the moles of NaOH, this number can be related to the moles of diprotic acid through the stoichiometric coefficients from the balanced equation, showing a fundamental relationship in stoichiometry between molarity, volume, and moles.