I need help filling out the rest of this sheet. it was for a lab i did but i don

Question

I need help filling out the rest of this sheet. it was for a lab i did but i dont know what to do. also please show work! thanks

Partner Mixture Volume in mL Section Calibration k Report values to 3 significant figures Hzo Absorbance LFeSCN on o2? Based upon the values for Kc, what can you conclude about the formula for the complex ion. Briefly explain. 2 200 253 000006 abs Slope Average Kc M 2.00E-03 Fe(NO)3 kscNE 2.00E-03 o oloL Initial Moles Mixture Equilibrium Moles Equilibrium Concentration SCN Kc O 00333 o ooo o 620 Mixture Initial Moles Equilibrium Moles Equilibrium Concentration Kc

Explanation / Answer

the first table is to form the calibration curve for standard solutions of known concentrations.

Acording to Beer's law,

Absorbance,A=elC ,where e=absorptivity,l=path length of light,C=concentration of absorbing ions.

this resembles the linear equation ,y=mx+c ,m=slope,c=intercept =0

m=slope=el

So if A vs C graph is plotted ,then Concentration =abs /slope...(1)

As slope is constant for a given ion species..

So for unknown solutions C can be calculated using equation (1)

Also ,Fe3+ +SCN- <-->FeSCN2+

kc=[FeSCN2+]eq/[Fe3+]eq [SCN-]eq

[FeSCN2+]eq=Abs/slope

[Fe3+]eq=[Fe3+]o-[FeSCN2+]eq

[SCN-]eq=[SCN-]o-[FeSCN2+]eq

eq subscript is used to denote equilibrium concentration,nd o subscript is used to denote initial concentration.

now coming back to your part 1 table,

i will assume that you have ploted the callibration curve using standard solution and found the slope to be 4082.9

Now in the part 1 table ,

trial 1)[Fe3+]o=5ml *2.0*10^-3M =0.005L *2.0*10^-3M=1.0*10^-5moles

[Fe3+]o=1.0*10^-5moles/total volume=1.0*10^-5moles/10ml=1.0*10^-5moles/0.01L=1.0*10^-3 M

[Fe3+]eq=[Fe3+]o-[FeSCN2+]eq

[FeSCN2+]eq=abs/slope=0.136/4082.9=0.0000333 M

[Fe3+]eq=1.0*10^-3 M-0.0000333 M=0.000967M

Similarly ,

[SCN]o=1ml *2.0*10^-3M =0.001L *2.0*10^-3M=2.0*10^-6moles

[SCN-]o=2.0*10^-6moles/0.01L=2.0*10^-4M

[SCN]eq=2.0*10^-4M-0.0000333 M=0.000167M

[SCN-]eq=0.000167M

kc=[FeSCN2+]eq/[Fe3+]eq [SCN-]eq=0.0000333 M/(0.000167M)(0.000967M)=206.206

kc=206.206 M-1

similarly calculate kc for al trials and take average ,by Kc=kc1+kc2+kc3.+kc4+kc5/5

kc has units M-1 so it is a second order reaction,the reactants must have reacted in 1:1 ratio to give the complex ion.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.