The reaction Fe3+(aq)+ SCN-(aq)FeSCN2+(aq)is initiated by mixing 1.50 mL of 0.0060 M iron nitrate (Fe3+) with...

The reaction Fe3+(aq)+ SCN-(aq)FeSCN2+(aq)is initiated by mixing 1.50 mL of 0.0060 M iron nitrate (Fe3+) with 0.75mL of 0.0020 M KSCN (SCN-) and enough water to give a total volume of 3.0 mL in a cuvette. The reaction goes to equilibrium with a final concentration of 1.55 x 10-4M FeSCN2+. Calculate Kc at 25°C for the reaction. Answer on the back side of this page. (Hint:calculate the initial concentrations of Fe3+ and SCN in the cuvette and use an ICE table

Expert Solution

concentration of Fe3+ = 1.50 x 0.0060 / 3.0

= 0.0030 M

concentration of SCN- = 0.75 x 0.0020 / 3.0

= 0.00050 M

Fe3+ (aq) + SCN- (aq) ------------> FeSCN2+ (aq)

0.0030 0.00050 M 0

0.0030-x 0.00050 - x x

[FeSCN2+] = x = 1.55 x 10^-4 M

[Fe3+] = 0.002845 M

[SCN-] = 3.45 x 10^-4 M

Kc = [FeSCN2+] / [Fe3+][SCN-]

= (1.55 x 10^-4) / (0.002845 )(3.45 x 10^-4 )

Kc = 158

queen_honey_blossom answered 2 years ago

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