Question

How many milliliters of 2.155 M KOH are required to titrate \(25.00 \mathrm{mL}\) of \(0.3057 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (prop-ionic acid)?

Short answer

3.544 milliliters of 2.155 M KOH are required to titrate 25 ml of 0.3057 M CH3CH2COOH (prop-ionic acid).

Step-by-step solution

Identify the Reaction

For every mole of CH3CH2COOH that reacts with KOH, it produces one mole of salt and one mole of water. So, the titration reaction is: CH3CH2COOH + KOH → CH3CH2COOK + H2O. According to this balanced reaction, the stoichiometric ratio of CH3CH2COOH to KOH is 1:1.

Calculate moles of CH3CH2COOH

We know the volume of CH3CH2COOH to be 25.00 ml or 0.025 liters and its molarity as 0.3057 M. Molarity is defined as the number of moles of solute per liter of solution, so we multiply the volume by the molarity to find the number of moles of CH3CH2COOH, which is \(0.025 L * 0.3057 M = 0.0076425 moles\)

Calculate the volume of KOH

Using the stoichiometry of the reaction (1:1), we know that we need an equal number of moles of KOH to titrate CH3CH2COOH. Given that the molarity of KOH is 2.155 M, we can find the required volume using the following formula: volume = moles / molarity. Therefore, the volume of KOH required in liters is \(0.0076425 moles / 2.155 M = 0.003544 liters\) or 3.544 milliliters.

Key concepts

Stoichiometry

Stoichiometry is the bridge between the quantities of reactants and products in a chemical reaction. It derives from the mole concept, which is a fundamental aspect of chemistry. In the context of our titration exercise, the stoichiometry of the reaction is provided by the equation:

• CH₃CH₂COOH + KOH → CH₃CH₂COOK + H₂O

Here, the stoichiometric ratio is 1:1, meaning one mole of propionic acid reacts with one mole of KOH. This simplifies calculations greatly, as the amount of acid directly equates to the amount of base needed for neutralization. Using this ratio helps determine how much of a reactant is needed based on the known quantity of another.
This principle is not only useful in laboratory settings but also plays a critical role in industrial applications where precise chemical compositions are necessary. Whether we're balancing equations or conducting experiments, stoichiometry allows chemists to predict the outcomes and proportions of substances involved in reactions.

Molarity

Molarity is a straightforward concept yet incredibly crucial in chemistry. It is defined as the number of moles of solute present in one liter of solution. This measurement helps chemists determine how concentrated a solution is.
In the titration problem, the propionic acid's molarity is given as 0.3057 M. This means there are 0.3057 moles of the acid in one liter of its solution. By knowing this concentration, we can calculate the exact amount of acid present in any volume of solution by multiplying the volume (in liters) by its molarity. For example, with 25.00 mL of propionic acid, we found \[0.025 \, L \times 0.3057 \, M = 0.0076425 \, ext{moles of CH}_3 ext{CH}_2 ext{COOH}\]Molarity is also used to find out how much of one substance is needed to react completely with another, which is critical in titration procedures where accuracy is key.

Acid-Base Reaction

Acid-base reactions are a staple of chemistry, describing the reactions between an acid (proton donor) and a base (proton acceptor). In this exercise, propionic acid ( CH₃CH₂COOH) acts as the acid and KOH as the base.
During the titration, KOH neutralizes the acid, forming water and potassium propionate (CH₃CH₂COOK). This type of reaction is specifically a neutralization reaction because it results in the neutralization of the acid and base properties to form a salt and water.

• Reaction: CH₃CH₂COOH + KOH → CH₃CH₂COOK + H₂O

By adding the KOH solution gradually to the acid solution, we can determine the point when all the acid has reacted, known as the equivalence point.
The concept of acid-base reactions is not only essential in chemistry education but also in numerous practical applications, ranging from food science to pharmaceuticals, showcasing the importance of understanding these reactions in everyday life.