In this lab, you will also carry out dilutions using a dropper to get qualitativ

Question

In this lab, you will also carry out dilutions using a dropper to get qualitative information about metal/ligand ratios. Determine the approximate answers to these questions: (a) Put two drops of hydroxylamine solution in a 0.50-mL well. Then add eight drops of 1.00 g L^-1 bipyridine and two drops of 5 mg L^-1 iron. How many moles of bipyridine and how many moles of iron are in the well? There are about 20 drops to 1 mL. (b) If you want to repeat the mixture from part (a) on a larger scale, describe how many milliliters of hydroxylamine solution, how many milliliters of bipyridine solution, and how many milliliters of iron solution you need to make 2.50 mL of total solution volume.

Explanation / Answer

(a). Amount of bipyridine added = 8 drops

Volume of bipyridine added = 8/20 mL = 0.0004 L

Concentration of bipyridine = 1.0 g/L

Mass of bipyridine = 0.1 * 0.0004 = 0.0004 g

Molar mass of bipyridine = 156.19 g/mole

Moles of bipyridine in well = 0.0004 / 156.19

= 2.56x10-6 moles

Amount of iron added = 2 drops

Volume of iron added = 2/20 mL = 0.0001 L

Concentration of iron = 5.0 mg/L

Mass of iron = 5.0 * 0.0001 = 0.0005 mg

Atomic mass of iron = 55.84 g/mole

Moles of iron in well = 0.0005 x 10-3 / 55.84

= 8.95x10-9 moles

(b). While preparing 0.5 mL solution, 0.4 mL bipyridine and 0.1 mL iron was taken.

Now we have to prepare 2.50 mL solution.

This means we have to prepare a solution with 5 times more volume.

So volume of bipyridine required = 5 * 0.4 = 2.0 mL

Volume of iron required = 5 * 0.1 = 0.5 mL

Also volume of hydroxylamine required = 5 * 0.1 = 0.5 mL

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