Question

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base were required for the titration?

Short answer

The concentration of acetic acid in the vinegar is 0.115 M.

Step-by-step solution

Calculate moles of Sodium Hydroxide

First, calculate the number of moles of the base, sodium hydroxide. We know from the question that the molarity (M) of NaOH is 1.00M, and the volume used in the titration is 5.75 mL (or 0.00575 L, because 1L = 1000 ml). The formula to calculate moles is: Moles = Volume(L) x Molarity(M). So, by substituting the values, we get the moles of NaOH = 1M x 0.00575 L = 0.00575 moles.

Find moles of Acetic Acid

Next, look at the stoichiometry of the reaction. In the balanced equation given, we see that one mole of NaOH reacts with one mole of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)). Hence, at the end point of the titration, the moles of NaOH that reacted should be equal to the moles of acetic acid in the vinegar solution. Therefore, we have 0.00575 moles of acetic acid present in our vinegar solution.

Determine concentration of Acetic Acid

Finally, calculate the molarity of the acetic acid. Remember that Molarity(M) = Moles/volume(L). From the question, we know that volume of the vinegar solution is 50.0 mL or 0.050 L. And from step 2,the moles of acetic acid found was 0.00575 moles. So substituting in the formula, we get Molarity of acetic acid: M = 0.00575 moles / 0.050 L = 0.115 M.

Key concepts

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's crucial in determining the quantities needed or formed in a reaction. In titration problems, stoichiometry helps us understand how substances interact in balanced chemical equations. Here, the balanced reaction between acetic acid \[\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\]shows a 1:1 mole ratio. This means every mole of acetic acid requires one mole of sodium hydroxide (NaOH) to react completely. This stoichiometric relation allows us to use the moles of NaOH used in the titration to find the moles of acetic acid present in the vinegar. Understanding stoichiometry is key to predicting the outcomes and quantities in any chemical reaction.

Molarity

Molarity is a measure of concentration. It tells you how much solute is dissolved in a given volume of solution. It is expressed in moles per liter (M). Knowing the molarity is essential in titration because it helps us find out how concentrated a solution is.
To find molarity, you divide the number of moles of solute by the volume of the solution in liters. In this exercise, we used this formula twice. First, to find the moles of sodium hydroxide using its known molarity and volume; then, to determine the molarity of acetic acid in the vinegar by using the moles obtained from the stoichiometry of the reaction. This helps understand how much acetic acid is in the vinegar, which is crucial for quality control in industries.

Chemical Reaction

A chemical reaction describes a process where substances (reactants) undergo a transformation to form new substances (products). In an acid-base titration, such as the one described, we observe how acetic acid reacts with sodium hydroxide. This reaction involves the transfer of protons from the acid to the base, resulting in the formation of sodium acetate and water.
By conducting the titration, we can find out how many moles of acetic acid were in the vinegar because we know how much of the base was needed to neutralize the acid. This is a classic example of a neutralization reaction, where an acid reacts with a base to form water and a salt, illustrating key concepts in both chemical reactions and stoichiometry. Acid-base reactions are integral in understanding chemistry and are widely applied in real-world situations, from laboratory experiments to industrial applications.