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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.

Short Answer

Expert verified
The molar mass of the diprotic acid can be determined by first finding the moles of NaOH in the given solution, using the volume and concentration: 0.1375 L × 0.750 M = 0.103125 mol. Then, use the stoichiometry to find the moles of the diprotic acid: 0.103125 mol ÷ 2 = 0.0515625 mol. Finally, calculate the molar mass of the diprotic acid by dividing the mass by the moles of the acid: 6.50 g / 0.0515625 mol ≈ 125.97 g/mol.

Step by step solution

01

Write the balanced chemical equation for the reaction

We can represent the diprotic acid as H₂A, where A is the conjugate base. The reaction between the diprotic acid and NaOH can be written as: \( H_2A + 2NaOH \rightarrow 2H_2O + Na_2A \) Since the acid is diprotic, one molecule of the acid will react with two molecules of NaOH.
02

Find moles of NaOH in the given solution

We are given the volume and concentration of the NaOH solution. From this, we can find the moles of NaOH in the solution: Moles of NaOH = Volume (in L) × Concentration Moles of NaOH = 0.1375 L × 0.750 M Moles of NaOH = 0.103125 mol
03

Use the stoichiometry to find moles of the diprotic acid

Using the balanced chemical equation, we know that 2 mol of NaOH reacts with 1 mol of the diprotic acid. Therefore, we can find the moles of the acid as follows: Moles of Acid = Moles of NaOH ÷ 2 Moles of Acid = 0.103125 mol ÷ 2 Moles of Acid = 0.0515625 mol
04

Calculate the molar mass of the diprotic acid

Now that we have the moles of the diprotic acid, we can find its molar mass by dividing the mass of the acid (given) by the moles of the acid: Molar mass of the acid = Mass / Moles of Acid Molar mass of the acid = 6.50 g / 0.0515625 mol Molar mass of the acid ≈ 125.97 g/mol Therefore, the molar mass of the diprotic acid is approximately \(125.97 \mathrm{~g/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
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.
Neutralization Reaction
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.}
Chemical Equation Balancing
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 and Moles
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.

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Most popular questions from this chapter

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