Question

From the data in the table, calculate the final [FeSCN²⁺]:

Beaker	0.200 M Fe(NO3)3 (mL)	0.0020 M SCN– (mL)	H2O (mL)	[FeSCN2+]
1	5.0	4.0	41.0
2	5.0	3.0	42.0
3	5.0	2.0	43.0
4	5.0	1.0	44.0

Answer 1

The reaction between Fe(NO₃)₃ and SCN^– to form FeSCN²⁺ is

Fe(NO₃)₃ + SCN^– -------> FeSCN²⁺

That means here all the components are euqimolar.i.e if we take 1M of Fe(NO₃)₃ and 1M of SCN^– , we will get 1 M of FeSCN²⁺

Now given

1- Beaker 1-

Initial Fe(NO₃)₃ taken = M1 = 0.200 M and V1 = 5.0 ml

Initial SCN^– taken = M1 = 0.0020 M and V1 = 4.0 ml

Water added = 41 ml

Now after the addition of these 3, the final volume = V2 = 5.0 ml +  4.0 ml + 41 ml = 50 ml

Thus the new concentration of Fe(NO₃)₃ in the total solution = M2 = M1V1/V2 = 0.200 M*5.0 ml / 50 ml = 0.02‬ M

Similarly new concentration of SCN^–  in the total solution = M2 = M1V1/V2 = 0.0020 M*4.0 ml / 50 ml = 0.00016‬ M

Thus in the final solution we have

0.02‬ M Fe(NO₃)₃ and 0.00016‬ M SCN^– . Since here SCN^– is present in less amount thus SCN^– is the limiting reagant.

Hence 0.00016‬ M SCN^– will react with 0.00016‬ M Fe(NO₃)₃ to form   0.00016‬ M FeSCN²⁺. The extra Fe(NO₃)₃ will remain as such in solution unreacted.

Thus [FeSCN²⁺] =  0.00016‬ M

2- Similarly-

Beaker 2-

Initial Fe(NO₃)₃ taken = M1 = 0.200 M and V1 = 5.0 ml

Initial SCN^– taken = M1 = 0.0020 M and V1 = 3.0 ml

Water added = 42 ml

Now after the addition of these 3, the final volume = V2 = 5.0 ml +  3.0 ml + 43 ml = 50 ml

Thus the new concentration of Fe(NO₃)₃ in the total solution = M2 = M1V1/V2 = 0.200 M*5.0 ml / 50 ml = 0.02‬ M

Similarly new concentration of SCN^–  in the total solution = M2 = M1V1/V2 = 0.0020 M*3.0 ml / 50 ml = 0.00012 M

Thus in the final solution we have

0.02‬ M Fe(NO₃)₃ and 0.00012‬ M SCN^– . Since here SCN^– is present in less amount thus SCN^– is the limiting reagant.

Hence 0.00012‬ M SCN^– will react with 0.00012‬ M Fe(NO₃)₃ to form   0.00012‬ M FeSCN²⁺. The extra Fe(NO₃)₃ will remain as such in solution unreacted.

Thus [FeSCN²⁺] =  0.00012 M

3-

Beaker 3-

Initial Fe(NO₃)₃ taken = M1 = 0.200 M and V1 = 5.0 ml

Initial SCN^– taken = M1 = 0.0020 M and V1 = 2.0 ml

Water added = 43 ml

Now after the addition of these 3, the final volume = V2 = 5.0 ml +  2.0 ml + 43 ml = 50 ml

Thus the new concentration of Fe(NO₃)₃ in the total solution = M2 = M1V1/V2 = 0.200 M*5.0 ml / 50 ml = 0.02‬ M

Similarly new concentration of SCN^–  in the total solution = M2 = M1V1/V2 = 0.0020 M*2.0 ml / 50 ml = 0.00008‬‬ M

Thus in the final solution we have

0.02‬ M Fe(NO₃)₃ and 0.00008‬‬ M SCN^– . Since here SCN^– is present in less amount thus SCN^– is the limiting reagant.

Hence 0.00008‬‬ M SCN^– will react with 0.00008‬‬ M Fe(NO₃)₃ to form   0.00008‬‬ M FeSCN²⁺. The extra Fe(NO₃)₃ will remain as such in solution unreacted.

Thus [FeSCN²⁺] =  0.00008‬‬ M

4-

Beaker 4-

Initial Fe(NO₃)₃ taken = M1 = 0.200 M and V1 = 5.0 ml

Initial SCN^– taken = M1 = 0.0020 M and V1 = 1.0 ml

Water added = 44 ml

Now after the addition of these 3, the final volume = V2 = 5.0 ml +  1.0 ml + 44 ml = 50 ml

Thus the new concentration of Fe(NO₃)₃ in the total solution = M2 = M1V1/V2 = 0.200 M*5.0 ml / 50 ml = 0.02‬ M

Similarly new concentration of SCN^–  in the total solution = M2 = M1V1/V2 = 0.0020 M*1.0 ml / 50 ml = 0.00004‬‬ M

Thus in the final solution we have

0.02‬ M Fe(NO₃)₃ and 0.00004‬‬ M SCN^– . Since here SCN^– is present in less amount thus SCN^– is the limiting reagant.

Hence 0.00004‬ M SCN^– will react with 0.00004‬‬ M Fe(NO₃)₃ to form   0.00004‬ M FeSCN²⁺. The extra Fe(NO₃)₃ will remain as such in solution unreacted.

Thus [FeSCN²⁺] =  0.00004‬ M