Calculate the concentrations of Fe and SCN- present in each of the beakers 1-5 a

Question

Calculate the concentrations of Fe and SCN- present in each of the beakers 1-5 as described in part I of the procedure. The total volume of the final mixture should be 25.0 mL in all cases. Note that the concentration of Fe3+ is much, much larger than the SCN concentration. This will cause the equilibrium position to lie far to the right, that is, nearly all of the SCN will react with Fe to form the complex ion and there will be large excess of unreacted Fe remaining in solution and very little SCN Calculate the concentration of FeSCN present in each beaker, assuming that all SCN has reacted to form the complex ion. These calculations must be done and written in your journal before coming to lab. 2+

Explanation / Answer

Answer:

Given:V

0.200 M of Fe(NO3)3 and 0.00180 M of KSCN.

Fe+3 + SCN- FeSCN+2

Thus Fe+3 and SCN- will combine 1:1 ratio initially.

Beaker-1: Concentration of Fe+3 = (Concentration of SCN-)* (Volume of SCN-)/Total volume of Solution

= 0.0018*1/25

=0.072*10-3 M

Beaker2 Concentration of Fe+3 = (0.0018*2)/25

= 0.144*10-3 M

Beaker3 Concentration of Fe+3 = (0.0018*3)/25

= 0.216*10-3 M

Beaker4 Concentration of Fe+3 = (0.0018*4)/25

= 0.288*10-3 M

Beaker5 Concentration of Fe+3 = (0.0018*5)/25

= 0.36*10-3 M

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