How much volume of a 1.5 M KOH is needed to extract a carboxylic acid from a mixture of 2.450 gram of benzoic acid and a 2.080 g of acetanilide? This is a liquid liquid extraction.
You need a distribution coefficient.
To determine the volume of a 1.5 M KOH solution needed for the liquid-liquid extraction of benzoic acid and acetanilide, we can follow these steps:
Step 1: Calculate the number of moles of benzoic acid and acetanilide.
To do this, divide the given masses of each compound by their respective molar masses. The molar mass of benzoic acid (C6H5COOH) is 122.12 g/mol, and the molar mass of acetanilide (C8H9NO) is 135.17 g/mol.
Number of moles of benzoic acid: 2.450 g / 122.12 g/mol = 0.020 mol
Number of moles of acetanilide: 2.080 g / 135.17 g/mol = 0.015 mol
Step 2: Determine which compound reacts with KOH.
In this case, benzoic acid (C6H5COOH) will react with KOH to form a water-soluble salt, while acetanilide (C8H9NO) will not.
Step 3: Use stoichiometry to calculate the required moles of KOH.
From the balanced equation:
C6H5COOH + KOH → C6H5COOK + H2O
The stoichiometry indicates that 1 mole of benzoic acid reacts with 1 mole of KOH. Therefore, we need the same number of moles of KOH as benzoic acid.
Required moles of KOH = 0.020 mol
Step 4: Calculate the volume of 1.5 M KOH solution.
The concentration of the KOH solution is given as 1.5 M, which means it contains 1.5 moles of KOH per liter of solution.
Volume of 1.5 M KOH solution = Required moles of KOH / Concentration of KOH solution
Volume of KOH solution = 0.020 mol / 1.5 mol/L ≈ 0.013 L
Converting liters to milliliters:
Volume of KOH solution = 0.013 L × 1000 mL/L = 13 mL
Therefore, approximately 13 mL of 1.5 M KOH solution is needed to extract the carboxylic acid from the given mixture.