Final answer:
To neutralize 0.490 g of crude aspirin containing 17.6% salicylic acid with 0.147 M NaOH, we require 4.25 mL of NaOH solution.
Step-by-step explanation:
To calculate the volume of 0.147 M NaOH required to neutralize 0.490 g of a crude aspirin sample containing 17.6% salicylic acid, we first determine the mass of the salicylic acid in the sample using the percentage purity given. The molecular weight of salicylic acid (C7H6O3) is 138.12 g/mol.
Calculating the mass of salicylic acid: 0.490 g × 17.6% = 0.08624 g of salicylic acid.
Next, we convert mass to moles: moles of salicylic acid = 0.08624 g ÷ 138.12 g/mol = 0.0006246 mol. The reaction between salicylic acid and NaOH is a 1:1 molar reaction, so 0.0006246 mol of NaOH is also needed to neutralize the salicylic acid.
To find the volume of NaOH, we use the molarity formula: Volume NaOH = moles NaOH ÷ molarity NaOH, which gives us Volume NaOH = 0.0006246 mol ÷ 0.147 M = 0.004248 mol/L. Therefore, to neutralize 0.490 g of crude aspirin, 4.25 mL (rounded to two decimal places) of 0.147 M NaOH is required.