Question

I'm having trouble filling out these charts, for the first chart (standard solutions) I found the mols for the concentration for SCN- so I am assuming it would be the same for FeNcs? For A.5 how do I figure out the percent transitmittance for each standard solution.

For the second chart the volume is the same for the test solutions 6-9 so would all of them have the same moles? For B4 I am guessing to just divide the ml if that test solution divided by 1000 multiplied by the concentration of the solution of SCN. And again in not really sure how to find the percent transmittance , and if you have an ideas on how I am supposed to graph this data that would be great!! Thank you

Experiment 34 An Equilibri ab Sec.Name Date_ 寸 이여 p 』

Answer 1

For the first table you are correct, moles of SCN:

$[SCN]=\frac{mols SCN}{\frac{V_f}{1000}}$ <-----Vf final volume 25mL

$[FeSCN^{2+}]=[SCN]$ <---- this is true assuming the following reaction:

$SCN^-+Fe^{3+}\rightarrow FeSCN^{2+}$

Transmittance is calcualted using

$%T=10^{2-A}$ $\%T=10^{(2-Abs)}$

	Blank	1	2	3	4	5
Vol NaSCN	0	1	2	3	4	5
moles SCN	0	0.000001	0.000002	0.000003	0.000004	0.000005
[SCN]	0	0.00004	0.00008	0.00012	0.00016	0.0002
FeNCS	0	0.00004	0.00008	0.00012	0.00016	0.0002
%T	100.000	77.625	57.016	42.364	33.963	29.040
Abs	0	0.11	0.244	0.373	0.469	0.537

Using this calibribation date you have to make a plot of Abs Vs [FeNCS]

And make a linear regression (straight line in the graph) to calculate the equation to calculate the concentration of FeNCS, thus the equation is:

$[FeNCS^{2+}]=0.0004(Abs)+0.000006$

For the second table:

The moles for Fe3+ are the same for all the solutions (you are adding the same volume)

$molsFe^{3+}=[Fe(NO_33)_3]*\frac{5}{1000}$

The moles of SCN :

$molsSCN^-=[NaSCN]*\frac{V_{(added)}}{1000}$ <----the volume added depending on the solution (1mL, 2mL, 3mL)

Trasnmittance the same as above.

	6	7	8	9	10
Vol Fe(NO3)3	5	5	5	5	5
moles Fe3+ (initial)	0.00001	0.00001	0.00001	0.00001	0.00001
Vol NaSCN	1	2	3	4	5
moles SCN	0.000002	0.000004	0.000006	0.000008	0.00001
%T	73.282	52.360	37.670	23.988	15.812
Abs	0.135	0.281	0.424	0.62	0.801

And using the equation from the first part you can calculate the equilibrium concentration of FeNCS each solution:

For example in 6:

$[FeNCS^{2+}]=0.0004(0.135)+0.000006=0.00006$

And the moles of FeNCS, you need to multiply the concentration by the volume of the solution (10mL)

$molFeNCS=[FeNCS^{2+}]*\frac{10}{1000}$

	6	7	8	9	10
[FeNCS] eq	0.00006	0.00012	0.00017	0.00025	0.00032
mols FeNCS	6.0E-07	1.2E-06	1.7E-06	2.5E-06	3.2E-06