Table 1 in the handout gives volumes of Fe3+ and SCN that you will use to make s
ID: 712138 • Letter: T
Question
Table 1 in the handout gives volumes of Fe3+ and SCN that you will use to make solutions to determine the molar absorptivity of FeSCN2+ . Create an organized, hand-written table in which you display the initial concentrations of Fe3+ and SCNand the estimated equilibrium concentration of FeSCN2+ for each solution. Be sure to include appropriate units. #
# Solution
Volume (ml) of 0.2M Fe(NO3)3 (in 0.10M HNO3)
2
# Solution
Volume (ml) of 0.2M Fe(NO3)3 (in 0.10M HNO3)
Volume (ml) of 0.002M KSCN (in 0.10M HNO3) Total volume (ml) 1 5.00 0.00 25.002
5.00 1.00 25.00 3 5.00 1.50 25.00 4 5.00 2.00 25.00 5 5.00 2.50 25.00Explanation / Answer
The equilibrium involved in the given complexation reaction :
Fe3+ (aq) +SCN-(aq) <---> FeSCN2+
As [SCN-]<<<[Fe3+] ,so it is assumed that the reaction is driven to completion as it is forward directed due to excess of one of the reactant .(le chatlier's principle).
The limiting reagent [SCN-] is completely used up ,and form equal concentration of product, FeSCN2+
[ FeSCN2+]= [SCN-] at equilibrium
solution # [Fe3+]initial =0.2M*Volume(in L)=(0.2mol/L*0.005L)/total volume=0.001 mol/0.025L=0.04M [SCN-]initial=0.002M*volume(in L) [ FeSCN2+]eqm= [SCN-]initial M 1) 0.04M =(0.002mol/L*0.0L)/0.025L=0 0 2) 0.04M =(0.002mol/L*0.001L)/0.025L=8.0*10^-5 M 8.0*10^-5 M 3) 0.04M =(0.002mol/L*0.0015L)/0.025L=1.2*10^-4 M 1.2*10^-4 M 4) 0.04M =(0.002mol/L*0.002L)/0.025L=1.6*10^-4 M 1.6*10^-4 M 5) 0.04M =(0.002mol/L*0.0025L)/0.025L=2.0*10^-4 M =2.0*10^-4 MRelated Questions
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