Of Rey from the following equilibrium concentrations FeSCN\": 1.71 × 10. M, Fe\"
ID: 590989 • Letter: O
Question
Of Rey from the following equilibrium concentrations FeSCN": 1.71 × 10. M, Fe": 8.28 × 10-4 M and [SCN]= 4.28 × 10-4 M In this experiment, you will prepare a standard solution for a calibration curve for the tormation of FeSCN complex by reaction different volumes of 0.002M Fe(NO,), and 0.002 M NaSCN. You will use 0.1M HNO, for all dilutions. Using the table, Calculate the equilibrium concentrations of FeSCN2 Vol of Fe Vol of SCN 0.1 M HNO3 Equilibrium (mL) 10.00 10.00 10.00 (mL) (mL) 10.00 9.00 8.00 7.00 6.00 concentration Fe[SCN 0.00 1.00 2.00 3.00 4.00 10.00 10.00 10.00 5.00 5.00 d Determination of Equilbrium Constant: Pre Lab 181 e Shatelier's Principle: Chemical Equilbrium anExplanation / Answer
5. Reaction,
Fe3+ + SCN- --> FeSCN2+
In all SCN- volume is lower than the Fe3+ volume (both have same molarity)
therefore, SCN- is the limiting reactant
Volume of SCN- (ml) Equilibrium FeSCN2+ (M)
0 0
1 0.002 M x 1 ml/20 ml = 0.0001
2 0.002 M x 2 ml/20 ml = 0.0002
3 0.002 M x 3 ml/20 ml = 0.0003
4 0.002 M x 4 ml/20 ml = 0.0004
5 0.002 M x 5 ml/20 ml = 0.0005
formula used,
molarity = moles/volume
moles = molarity x volume
final molarity = initial molarity x initial volume/total volume
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