10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3 M KSCN and...

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10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3M KSCN and 6.00 mL of water. The equalibrium molarity of FeSCN2+ isfound to be 8.70E-5 M. If the reaction proceeds as shown below,what is the equilibrium molarity of Fe3+ and SCN-? Fe3+(aq) +SCN-(aq) FeSCN2+(aq)
Equilibrium Concentration Fe3+
Equilibrium Concentration SCN-

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4.4 Ratings (1097 Votes)

initial molarity of Fe+3 = 10 x 3.48 x 10^-3 / (10 +4 + 6)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.74 x 10^-3 M

initial molarity of SCN- = 4 x 2.84 x 10^-3 / 20

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 5.68 x 10^-4 M

Fe+3Â Â Â Â Â Â Â Â Â Â Â +Â Â Â Â SCN-Â Â Â ----------------------------> FeSCN2+

1.74 x 10^-3Â Â Â Â Â 5.68 x 10^-4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ----------------> initial

1.74 x 10^-3-xÂ Â 5.68 x 10^-4 -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -------------------> equilibrium

x = 8.70 x 10^-5 M

equilibrium concentration of Fe+3 = 1.74 x 10^-3-x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.74 x 10^-3 - 8.70 x 10^-5

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.65 x 10^-3 M

equilibrium concentration of SCN- = 5.68 x 10^-4 -xÂ Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =5.68 x 10^-4 - 8.70 x 10^-5

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.81 x 10^-4 M

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10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3 M KSCN and...

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