can someone do questions 4, 6 & 8 please and thank you! 3. Record the concentrat
ID: 572712 • Letter: C
Question
can someone do questions 4, 6 & 8 please and thank you!
3. Record the concentrations of the following stock solutions as posted on Desire2Learn (D2L); stock solution containing Feions: Fe(NOs)s solution2.5 E-3M stock solution containing SCN ions: NH SCN solution2. SIE-2M Calculate the concentration of the dilute Fe3+ solution which will be prepared in Part I step 2 of the procedure. Use the exact concentration of the stock solution posted on Desire2Learn (D2L). 4. v-tml V .oot Chemistry 213-Winter 2018Explanation / Answer
4) from part 1) It can be seen that the stock solution is used up for preparing various standard solutions of varying concentrations.
Concentration of stock solution of Fe3+=2.5E-3=C(stock)
volume of stock solution used=4ml=0.004L
Concentration of dilute solution of Fe3+=C(dil)
volume of dil solution=1L [you have prepared 1L of dilute solution)
Using equation,
C(dil)*V(dil)=C(stock)*V(stock)
C(dil)=C(stock)*V(stock)/V(dil)=(2.5*10^-3 mol/L)*0.004L/1L=0.01*10^-3 mol/L=1*10^-5mol/L=1E-5 mol/L
6) sample calculation for test tube1:
Dil Fe(NO3)3=0.001L
mol of Fe3+ ions added=C(dil)*0.001L=1*10^-5mol/L*0.001L=1*10^-8mol
Due to large excess of SCN- ions ,all the Fe3+ ions get converted to Fe(SCN)2-
Fe3+ +SCN-<--->Fe(SCN)2-
So, mol of Fe(SCN)2- ions formed=1*10^-8mol
[Fe(SCN)2-]=[Fe3+]=1*10^-8mol/total volume of solution=1*10^-8mol/10ml=1*10^-8mol/0.01L=1*10^-6 mol/L
[Fe(SCN)2-]=1*10^-6 mol/L=equilibrium concentration
8)Plot a callibration curve between Absorbance vs concentration of Fe3+ solution from table 1(for standard solution)
Absorbance=eCl [Beer's law]
where e=molar absorptivity of Fe3+
l=path length of light
el=slope
Thus find the slope from the callibration curve ,and use the value to cakculate concentration of unknown Fe3+ solutions of known absorbance using eqn, C(unknown)=Absorbance(unknown)/el
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