A SOLUTION IS MADE BY MIXING 5.00 mL .00300 MFE(NO3) WITH 4.00mL 0.00300 M KSCN

Question

A SOLUTION IS MADE BY MIXING 5.00 mL .00300 MFE(NO3) WITH 4.00mL 0.00300 M KSCN AND 3.00 mL 1.0 M HNO3. AFTER EQUILIBRIUM ISESTABLISHED, THE CONCENTRATION OF Fe(SCN)+2 WAS DETERMINED TO BE2.72*10E-4 M. CALCULATE THE EQUILIBRIUM CONSTANT FOR THISREACTION. Fe=3 (aq) + SCN- (aq) <-----> Fe(SCN)+2 (aq) A SOLUTION IS MADE BY MIXING 5.00 mL .00300 MFE(NO3) WITH 4.00mL 0.00300 M KSCN AND 3.00 mL 1.0 M HNO3. AFTER EQUILIBRIUM ISESTABLISHED, THE CONCENTRATION OF Fe(SCN)+2 WAS DETERMINED TO BE2.72*10E-4 M. CALCULATE THE EQUILIBRIUM CONSTANT FOR THISREACTION. Fe=3 (aq) + SCN- (aq) <-----> Fe(SCN)+2 (aq)

Explanation / Answer

Total initial volume 5 + 4 + 3 = 12 mL = 0.012 L . Moles of Fe3+ = molarity * volume (L)                        =0.003 M * 0.005 L                         =0.000015 moles . Initial concentration = moles / volume                              = 0.000015 moles/ 0.012 L                               =0.00125 M . Moles of SCN- = 0.003 M * 0.004 L                         =0.000012 moles . Initial concentration = 0.000012 moles / 0.012 L                               =0.001 M .                   Fe+3(aq) + SCN- (aq) <-----> Fe(SCN)+2(aq) . The equilibrium concentration of Fe(SCN)2+ wasfound to be 2.72 * 10^-4 M Since the initial concentration of this species in zero, thereduction in concentration of reactents is equal to the increase inconcentration of this species i.e. 2.72 * 10^-4 M . Hence equilibrium concentration of Fe3+ = 0.00125 M - (2.72 *10^-4 M)                                                                =0.000978 M . Equilibrium concentration of SCN- = 0.001 M - (2.72 * 10^-4M)                                                      =0.000728 M . Equilibrium constant, Kc = [Fe(SCN)+2] / {[ Fe+3] [ SCN-] }                                       =(2.72 * 10^-4 M) / { 0.000978 M * 0.000728 M }                                        =318.82                   Fe+3(aq) + SCN- (aq) <-----> Fe(SCN)+2(aq) . The equilibrium concentration of Fe(SCN)2+ wasfound to be 2.72 * 10^-4 M Since the initial concentration of this species in zero, thereduction in concentration of reactents is equal to the increase inconcentration of this species i.e. 2.72 * 10^-4 M . Hence equilibrium concentration of Fe3+ = 0.00125 M - (2.72 *10^-4 M)                                                                =0.000978 M . Equilibrium concentration of SCN- = 0.001 M - (2.72 * 10^-4M)                                                      =0.000728 M . Equilibrium constant, Kc = [Fe(SCN)+2] / {[ Fe+3] [ SCN-] }                                       =(2.72 * 10^-4 M) / { 0.000978 M * 0.000728 M }                                        =318.82

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