Use malarity and titration calculations
To illustrate the concept of molarity and titration calculations, let's consider a hypothetical scenario where we are trying to determine the concentration of acetic acid (CH3COOH) in a vinegar sample.
First, we need to prepare a standardized solution of sodium hydroxide (NaOH) of known concentration. Let's say we prepare a 0.1M NaOH solution.
Next, we perform a titration by adding the NaOH solution to the vinegar sample until the equivalence point is reached. The balanced chemical equation for the reaction between acetic acid and sodium hydroxide is:
CH3COOH + NaOH → CH3COONa + H2O
We can use the stoichiometry of the reaction to determine the molarity of acetic acid in the vinegar sample. Let's say it took 25 ml of the NaOH solution to reach the equivalence point with 30 ml of vinegar.
From the titration, we can determine that it took 25 ml of 0.1M NaOH solution to neutralize the acetic acid in 30 ml of vinegar. Using this information, we can calculate the molarity of acetic acid in the vinegar sample:
Molarity of NaOH = 0.1 M
Volume of NaOH = 25 ml = 0.025 L
Moles of NaOH = Molarity x Volume = 0.1 M x 0.025 L = 0.0025 moles
According to the balanced chemical equation, 1 mole of acetic acid reacts with 1 mole of NaOH. Therefore, the moles of acetic acid in the vinegar sample is also 0.0025 moles.
Volume of vinegar = 30 ml = 0.03 L
Molarity of acetic acid = Moles of acetic acid / Volume of vinegar = 0.0025 moles / 0.03 L = 0.0833 M
Therefore, the molarity of acetic acid in the vinegar sample is 0.0833 M. This calculation demonstrates the use of molarity and titration in determining the concentration of a substance in a solution.