A 1.00 mL aliquot of solution A was diluted to 50.00 mL using 0.02 M Fe(III), forming solution B (reaction 2)
By looking at this how many moles of sodium salicylate were in the 1.00 mL aliquot used to prepare solution B?
DrBob222, not helpful at all.
To determine the number of moles of sodium salicylate in the 1.00 mL aliquot used to prepare solution B, we need to use the given information and some stoichiometry.
First, let's understand the reaction that takes place between solution A and the 0.02 M Fe(III):
Fe(III) + Sodium Salicylate -> Complex
From the given information, we know that 1.00 mL of solution A was diluted to 50.00 mL using the 0.02 M Fe(III), resulting in solution B. This dilution process indicates that the concentration of sodium salicylate in solution A is the same as the concentration of Fe(III) in solution B (since they react in a 1:1 ratio).
To calculate the number of moles of sodium salicylate, we need to determine the number of moles of Fe(III) in solution B. This can be found using the following formula:
Moles of Fe(III) = Concentration of Fe(III) × Volume of Solution B
Given that the concentration of Fe(III) is 0.02 M and the volume of solution B is 50.00 mL (or 0.0500 L), we can calculate the moles of Fe(III):
Moles of Fe(III) = 0.02 M × 0.0500 L = 0.00100 moles
Since the reaction between Fe(III) and sodium salicylate occurs in a 1:1 ratio, the number of moles of sodium salicylate in the 1.00 mL aliquot used to prepare solution B is also 0.00100 moles.