At what approximate wavelength range (in nm) would you expect a red/purple solut
ID: 489153 • Letter: A
Question
At what approximate wavelength range (in nm) would you expect a red/purple solution to have a maximum? Knowing the molecular weight of the salicylic acid (MW = 138.12 g/mol), calculate the amount of salicylic acid (in grams) that is needed to prepare 100-mL of a 5.00 Times 10^-3M salicylic acid volumetric solution. The volume ratio between stock salicylic acid and stock ferric chloride should be 5:1 (i.e. 5 parts of salicylic acid to 1 part of ferric chloride) for the solution. You will add enough distilled water to fill the solution to the desired total volume. Based on the above instruction, what volumes of 2.50 Times 10^-3 M salicylic acid solution and 0.15 M ferric chloride solution would you need in order to prepare 50-mL of iron(III)-salicylate volumetric solution that has a concentration of 5.00 Times 10"4 M in the iron(III)-salicylate complex? Report the volumes to two decimal places. Refer to reaction #3 shown on page 5 of the procedure for the reaction between salicylic acid and ferric chloride.Explanation / Answer
2) Calculate the moles of salicylic acid present in 100 mL of 5.00*10-3 M salicylic acid solution = (100 mL)*(1 L/1000 mL)*(5.00*10-3 mol/L) = 0.0005 mole.
Calculate the mass of salicylic acid required = (moles of salicylic acid required)*(molecular weight of salicylic acid) = (0.0005 mole)*(138.12 g/mol) = 0.06906 g 0.070 g (ans).
3) Write down the balanced chemical equation between Fe(III) and salicylic acid (SA) as
Fe3+ + 3 SA --------> [Fe(SA)3]
As per the balanced stoichiometric equation,
1 mole Fe3+ = 3 moles SA = 1 mole iron (III) salicylate.
We have 50 mL of 5.00*10-4 M iron (III) salicylate; therefore, moles of iron (III) salicylate = (50 mL)*(1 L/1000 mL)*(5.00*10-4 mol/L) = 2.5*10-5 mole.
Moles of SA required = (2.5*10-5 mole iron (III) salicylate)*(3 moles SA/1 mole iron (III) salicylate) = 7.5*10-5 mole.
Molar concentration of stock SA = 2.50*10-3M; therefore, volume of SA required = (7.5*10-5 mole)/(2.5*10-3 mol/L) = 0.03 L = (0.03 L)*(1000 mL/1 L) = 30 mL.
Again, 1 mole Fe3+ = 1 mole iron (III) salicylate.
Therefore, moles of Fe3+ required = (2.5*10-5 mole iron (III) salicylate)*(1 mole Fe3+/1 mole iron (III) salicylate) = 2.5*10-5 mole.
Molar concentration of FeCl3 = 0.15 M.
Therefore, volume of FeCl3 required = (2.5*10-5 mole)/(0.15 mol/L) = 1.667*10-4 L = (1.667*10-4 L)*(1000 mL/1 L) = 0.1667 mL 0.17 mL (ans).
We see that we require a much lower volume of iron chloride and hence, iron chloride is the limiting reactant. The volume of SA required will be (0.17*5) mL = 0.85 mL (ans).
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