Question

An artificial fruit beverage contains \(12.0 \mathrm{~g}\) of tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6},\) to achieve tartness. It is titrated with a basic solution that has a density of \(1.045 \mathrm{~g} / \mathrm{cm}^{3}\) and contains 5.00 mass percent KOH. What volume of the basic solution is required? (One mole of tartaric acid reacts with two moles of hydroxide ion.)

Short answer

Answer: The required volume of the KOH solution to titrate 12.0 g of tartaric acid is 171.812 ml.

Step-by-step solution

Calculate the moles of tartaric acid

First, we convert the given mass of tartaric acid (\(12.0 \mathrm{~g}\)) to moles by dividing by the molar mass of tartaric acid (\(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\)). The molar mass of tartaric acid is approximately \(150.087 \mathrm{~g/mol}\). Therefore, the moles of tartaric acid are: Moles\(_{\text{Tartaric Acid}}\) = \(\frac{12.0 \text{ g}}{150.087 \text{ g/mol}}\) = \(0.08 \text{ mol}\)

Determine the moles of KOH required

As mentioned in the exercise, one mole of tartaric acid reacts with two moles of hydroxide ion. Therefore, based on the stoichiometry, the moles of KOH required can be calculated as follows: Moles\(_{\text{KOH}}\) = \(0.08 \text{ mol} \times 2\) = \(0.16 \text{ mol}\)

Find the mass of KOH needed

Now that we have found the moles of KOH required, we can find the mass of KOH needed by multiplying the moles with the molar mass of KOH (\(56.105 \mathrm{~g/mol}\)): Mass\(_{\text{KOH}}\) = \(0.16 \text{ mol} \times 56.105 \mathrm{~g/mol}\) = \(8.9776 \text{ g}\)

Calculate the volume of KOH solution required

In the given KOH solution, there is \(5.00\%\) mass of KOH. To find the mass of the solution that corresponds to the required mass of KOH, we can use the formula: Mass\(_{\text{solution}}\) = \(\frac{\text{Mass}_\text{KOH}}{\% \text{ mass KOH}}\) = \(\frac{8.9776 \text{ g}}{0.0500}\) = \(179.552 \text{ g}\) Next, we use the density of the solution to find the volume required: Volume\(_{\text{solution}}\) = \(\frac{\text{Mass}_\text{solution}}{\text{Density}} = \frac{179.552 \text{ g}}{1.045 \mathrm{~g/cm}^3}\) = \(171.812 \mathrm{~cm}^3 = 171.812 \mathrm{~ml}\) The required volume of the basic KOH solution is \(171.812 \mathrm{~ml}\).

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced during a reaction, given the balanced chemical equation. For example, if we consider the acid-base reaction between tartaric acid, \(\mathrm{H}_2 \mathrm{C}_4 \mathrm{H}_4 \mathrm{O}_6\), and potassium hydroxide (KOH), stoichiometry tells us that it takes two KOH molecules to neutralize one molecule of tartaric acid. Therefore, to calculate the required amount of KOH to neutralize a certain mass of tartaric acid, we need to know the stoichiometric ratio, which is 1:2 in this case.

Understanding the stoichiometric relationship helps in titration calculations as well, where the goal is to precisely measure how much of a titrant (like KOH in our example) is needed to react completely with an analyte (such as tartaric acid).

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It can be calculated by summing the atomic masses of all the atoms in a molecule. The molar mass is essential in converting between moles and grams, a common step in stoichiometry.

For instance, the molar mass of tartaric acid, \(\mathrm{H}_2 \mathrm{C}_4 \mathrm{H}_4 \mathrm{O}_6\), is approximately \(150.087 \text{ g/mol}\). This means that one mole of tartaric acid weighs about 150.087 grams. By knowing this value, we can determine the number of moles of tartaric acid in a 12.0-gram sample, which enables us to carry out subsequent titration calculations.

Acid-Base Reaction

An acid-base reaction is a chemical process in which an acid reacts with a base to produce water and a salt. In the context of a titration, the base (titrant) is added to the acid (analyte) until the reaction reaches its equivalence point, where the number of moles of hydrogen ions equals the number of moles of hydroxide ions.

In our example, tartaric acid acts as the acid and KOH acts as the base. The stoichiometry of the reaction dictates that two hydroxide ions from KOH are needed to react with one tartaric acid molecule. Understanding the nature of this reaction is key to determining the end point of the titration and, consequently, the volume of titrant needed.

Solution Concentration

Solution concentration refers to the amount of solute (such as KOH) present in a given quantity of solvent or solution. It can be expressed in various ways, including percent by mass, molarity, molality, or normality. In titrations, knowing the concentration of the titrant is crucial for accurate calculations. For our example, we are given a KOH solution with a concentration of 5.00 mass percent, meaning there are 5 grams of KOH in every 100 grams of solution.

Once we determine the mass of KOH needed to react with the tartaric acid using stoichiometry and molar masses, we can then calculate the mass of the KOH solution required by considering its concentration. We then convert this mass to volume using the density of the KOH solution, completing the titration calculation.