Question

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

Short answer

The molar mass of the diprotic acid is 126.02 \(\mathrm{g/mol}\).

Step-by-step solution

Write the Balanced Equation for the Reaction

Let's call our diprotic acid H2A. Then the balanced equation for the reaction between NaOH and the diprotic acid is: H2A + 2NaOH -> 2H2O + Na2A Note that a diprotic acid has two acidic protons, so the stoichiometry is 1 mol of the acid reacts with 2 mol of NaOH.

Calculate the Amount of NaOH in Moles

We need to find out how many moles of NaOH were used in the reaction. We are given the volume and concentration of the NaOH solution, so we can use the formula: Moles of NaOH = Volume x Concentration Moles of NaOH = \(137.5\mathrm{~mL}\) × 0.750 M Note that we need to convert the volume of NaOH into liters so that the units are matched: Moles of NaOH = \(0.1375\mathrm{~L}\) × 0.750 M = 0.103125 mol

Calculate the Amount of the Acid in Moles

Next, we need to convert the moles of NaOH to moles of the diprotic acid using the stoichiometric coefficients in the balanced equation. We can use the ratio of moles of H2A to moles of NaOH: Moles of H2A = Moles of NaOH × \(\frac{1 \ \mathrm{mol\ H2A}}{2 \ \mathrm{mol\ NaOH}}\) Moles of H2A = \(0.103125\mathrm{~mol}\) × \(\frac{1 \ \mathrm{mol\ H2A}}{2 \ \mathrm{mol\ NaOH}}\) = 0.0515625 mol

Calculate the Molar Mass of the Acid

Now that we have the moles of H2A, we can use the given mass of the sample to find the molar mass, using the formula: Molar Mass = \(\frac{\mathrm{Mass\ of\ a\ Sample}}{\mathrm{Moles}}\) Molar Mass = \(\frac{6.50\mathrm{~g}}{0.0515625 \mathrm{~mol}}\) = 126.02 \(\mathrm{g/mol}\) So, the molar mass of the diprotic acid is 126.02 \(\mathrm{g/mol}\).

Key concepts

Diprotic Acid

A diprotic acid is a type of acid that can donate two protons or hydrogen ions ( +H^+) per molecule in a reaction. This unique characteristic makes it differ from monoprotic acids, which only donate one proton.
When a diprotic acid is involved in a chemical reaction, it generally reacts in two distinct stages, each releasing one proton at a time.
In the case of the reaction between a diprotic acid (such as H2A) and a strong base like NaOH, the stoichiometry becomes crucial. This is because for every one molecule of diprotic acid, you need two molecules of the strong base to achieve full neutralization.
Diprotic acids are important in various fields, especially in titration and stoichiometry, as they require careful consideration of their two-stage proton donation. Understanding the nature of diprotic acids helps in calculating moles accurately when dealing with neutralization reactions.

Stoichiometry

Stoichiometry is a branch of chemistry that involves the calculation of reactants and products in chemical reactions. It uses the principles of the law of conservation of mass to balance chemical equations.
In a chemical reaction, stoichiometry helps us understand the relationship between the quantities of the reactants and the products. It is represented by the coefficients in a balanced chemical equation.

• These coefficients indicate the ratio of moles of each substance involved in the reaction.
• For example, in a reaction involving a diprotic acid (H2A) and NaOH, the balanced equation is: H2A + 2NaOH → 2H2O + Na2A.

This shows a 1:2 stoichiometry between the diprotic acid and NaOH. Therefore, to fully neutralize the acid, twice as many moles of NaOH are needed compared to the moles of the diprotic acid. This molar ratio is key in calculating the unknowns in a chemical reaction, like determining the amount of reactant needed or product formed.

Neutralization Reaction

A neutralization reaction is a chemical reaction in which an acid and a base react to form water and a salt, resulting in the balancing of the acidic and basic properties of the substances involved.
This reaction occurs when the hydrogen ion ( +H) and the hydroxide ion ( −OH^−) combine to form water ( H_2O), thus neutralizing the acidic and basic components.
In the context of a diprotic acid and NaOH, the neutralization process involves the consumption of two equivalents of NaOH for one equivalent of the diprotic acid, as seen in the reaction: H2A + 2NaOH → 2H2O + Na2A.

• Neutralization reactions are essential in titration processes, where the concentration of an unknown solution can be determined by reacting it with a solution of known concentration.

Understanding neutralization is critical to setting up and solving equations related to stoichiometry and other calculations in chemistry.

Moles Calculation

Moles calculation is fundamental in chemistry to quantify the amount of substance in a reaction. The concept of the mole connects the mass of particles to the number of particles themselves via Avogadro's number.
In the reaction between a diprotic acid and NaOH, calculating the moles of NaOH used is the first step. This is done by multiplying the concentration of NaOH (M) by the volume (L), ensuring that volumes are converted to liters when needed:\[ \text{Moles of NaOH} = 137.5 \text{ mL} \times 0.750 \text{ M} = 0.103125 \text{ mol} \]
Once you have the moles of NaOH, you can find the moles of the diprotic acid using the stoichiometric relationship described in the balanced equation:\[ \text{Moles of H2A} = \text{Moles of NaOH} \times \frac{1}{2} = 0.0515625 \text{ mol} \]This approach is crucial to solve many chemistry problems, linking laboratory measurements with theoretical quantities necessary for precise calculations.