Determination of K C for the reaction: Fe 3+ (aq) + SCN - (aq) --> [Fe (SCN)] 2+

Question

Determination of KC for the reaction: Fe3+ (aq) + SCN- (aq) --> [Fe (SCN)] 2+ (aq)

using the average value for KC and the concentrations provided for the reactants, determine the absobance expected when 16.00 mL of Fe3+ (aq) is reacted with 10 mL of SCN (aq) and measured in a 1.00 cm cuvette with 2200 L/(mol.cm) molar absorptivity(E) for [Fe (SCN)]2+ at a wavelength of 447nm. (hint: need to solve quadratic equation and use beer`s law to solve absorbance)

average value of KC = 1.07 x 103

concentration of Fe3+ = 0.00111 mol/L

concentration of SCN = 0.00111 mol/L

beer`s law is A= E.b.c

E= 2200 L/(mol.cm) b= 1.00cm

Explanation / Answer

Determination of KC for the reaction: Fe3+ (aq) + SCN- (aq) --> [Fe (SCN)] 2+ (aq)

using the average value for KC and the concentrations provided for the reactants, determine the absobance expected when 16.00 mL of Fe3+ (aq) is reacted with 10 mL of SCN (aq) and measured in a 1.00 cm cuvette with 2200 L/(mol.cm) molar absorptivity(E) for [Fe (SCN)]2+ at a wavelength of 447nm. (hint: need to solve quadratic equation and use beer`s law to solve absorbance)

average value of KC = 1.07 x 103

concentration of Fe3+ = 0.00111 mol/L

concentration of SCN = 0.00111 mol/L

beer`s law is A= E.b.c

E= 2200 L/(mol.cm) b= 1.00cm

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ANS

moles of Fe3+=volume*molarity= 0.00111 mol/L*0.016L=0.00001776moles

moles of SCN=volume*molarity= 0.00111 mol/L*0.01L=0.0000111moles

molarity after mixing two solution

[Fe3+]=0.00001776moles/0.026L=0.000683M

[SCN]=0.0000111moles/0.026L=0.000427M

since SCN is limiting reagent the effective initial concentrations are

[Fe3+]=[SCN]=0.000427M

kc=1.07 x 103=[Fe (SCN)] 2+]/{[Fe3+][SCN]}

      Fe3+ (aq) + SCN- (aq) --> [Fe (SCN)] 2+ (aq)

0.000427........0.000427

-x.........................-x.......................+x

0.000427-x.......0.000427-x....................x

kc=1070=[Fe (SCN)] 2+]/{[Fe3+][SCN]}=x/{(0.000427-x)(0.000427-x)}

{1070{(0.000427)2 -2x(0.000427)+x2}=x

x2-1.000854x+0.0001951=0

x=c=[SCN]=0.000195M equillibrium concentration of SCN

A= E.b.c

E= 2200 L/(mol.cm)

b= 1.00cm

so A=2200 L/(mol.cm)*1.00cm*0.000195moles/L

A=0.429 (ANS)

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