Given, 5 100mL flasks each filled with 5.00mL of 0.00139 M ligand solution, 5.00mL of 2M...

Given, 5 100mL flasks each filled with 5.00mL of 0.00139 M ligand solution, 5.00mL of 2M sodium acetate, and 4.00mL of 3M NH_2OH. Then 1.00mL, 2.00mL, 3.00mL, 4.00mL and 5.00mL of 0.255 M Fe2+ is added repectivly to each flask. After that, each flask is filled to the 100mL mark with distilled water. What is the [Fe^2+] after dilution for each flask? given in M. Please show work, thank you.

Expert Solution

In flask 1, only 1 mL of 0.255 M Fe2+ is added, rest is other reagents. So dilution factor = 100/1 = 100

So, final conc after dilution = 0.255/100 = 0.00255 M

In flask 2, only 2 mL of 0.255 M Fe2+ is added, rest is other reagents. So dilution factor = 100/2 = 50

So, final conc after dilution = 0.255/50 = 0.00510 M

In flask 3, only 3 mL of 0.255 M Fe2+ is added, rest is other reagents. So dilution factor = 100/3

So, final conc after dilution = 0.255/(100/3) = 0.00765 M

In flask 4, only 4 mL of 0.255 M Fe2+ is added, rest is other reagents. So dilution factor = 100/4 = 25

So, final conc after dilution = 0.255/25 = 0.0102 M

In flask 5, only 5 mL of 0.255 M Fe2+ is added, rest is other reagents. So dilution factor = 100/5 = 20

So, final conc after dilution = 0.255/20 = 0.01275 M

queen_honey_blossom answered 2 years ago

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