I\'m trying to fill out the data sheet for my homework; initial moles, equilibri
ID: 1087329 • Letter: I
Question
I'm trying to fill out the data sheet for my homework; initial moles, equilibrium moles, equilibrium concentrations, and Kc (equilibrium constant) but was given different information in class as opposed to what my homework says to do and now I am so confused. Can someone PLEASE help me out?
UPDATED
Experimental Determination of the calibration curve (to be performed in pairs) By preparing solutions with a small amount of HSCN and a large excess of Fe", complete conversion of HSCN into FeSCN2 is effectively assured, and the concentration of FeSCN2 can be determined Using a 100 ml volumetric flask, prepare four soluti 8.0 ml of 2.00 x 103 M in the four resulting solutions will be as follows ons of FeSCN by diluting 2 0,40,6.0, and HSCN to 100 ml with 0.200 M Fe(NO,)>. The concentrations of FesCN2 Dil uti ng 2.0 ml 2.00 x 103 M HSCN gives a solution that has [FeSCN21 = 0.40 x 104 M . Diluting 4.0 ml 2.00 x 103 M HSCN gives a solution that has [FeSCN-]-080 x 104 M . Diluting 6.0 ml 2.00 x 10° M HSCN gives a solution that has [FeSCN21 1.2 x 10 M . Diluting 8.0 ml 2.00x 103 M HSCN gives a solution that has [FeSCN2] 1.6x 10 M Measure the absorbance of these solutions at 447 nm, using 0.200 M Fe(NO), as a reference solution. Record these absorbances on the data sheet and construct a graph of absorbance vs [FesCN2], and draw the best straight line passing through the origin.Explanation / Answer
Information given in your class only needs to be refered to solve the Homework data table.
Information useful:
Firstly draw the callibration curve of Absorbance Vs [FeSCN2+] (for standard solutions) using the callibration curve data table.
From this callibration curve, you can get the value of [FeSCN2+] for your test solutions ,solutions 1-5 ,
table for determination of Equilibrium constant (solved for solution 1 data)
calibration curve equation,y=4487.5x+0.0495
x=(y-0.0495)/4487.5 [y=Abs,x=[FeSCN2+]
x=(0.087-0.0495)/4487.5=8.356*10^-6 M
[H+] eq=[FeSCN2+]eq=8.356*10^-6M
calculation of initial moles
test tube 1
[Fe3+]o=2.0*10^-3 mol/L *5ml=2.0*10^-3 mol/L *0.005 L=1.0*10^-5 mol
[HSCN]=2.0*10^-3 mol/L *1ml=2.0*10^-3 mol/L *0.001 L=2.0*10^-6 mol
calculation of equilibrium moles:
(FeSCN2+)eq or eqm mol=eqm molarity *total volume=8.356*10^-6 mol/L *0.010 L=8.356*10^-8 mol
(Fe3+ )eq=(Fe)o-(FeSCN2+)eq =(1.0*10^-5 mol)-(8.356*10^-8 mol)=(1.0*10^-5 mol)-(0.008356*10^-5 mol)=0.992*10^-5M
equilibrium concentration=eqm mol/total volume=0.992*10^-5 mol/0.010L=0.992*10^-3 mol/L
(HSCN)eq=(HSCN)o-(FeSCN2+)eq=2.0*10^-6 mol-(8.356*10^-8 mol)=2.0*10^-6 mol-(0.008356*10^-5 mol)=1.992*10^-5 mol
equilibrium concentration=eqm mol/total volume=1.992*10^-5 mol/0.010L=1.992*10^-3 mol/L
kc=[FeSCN2+][H+]/[Fe3+][HSCN]=(8.356*10^-6M)(8.356*10^-6M)/(0.992*10^-3 mol/L)(1.992*10^-3 mol/L)=35.334*10^-6=3.533*10^-5
Equilibrium concentrations solution Absorbance Fe3+ HSCN FeSCN2+ H+ 1 0.087 8.356*10^-6M 8.356*10^-6M