A solution was prepared by mixing 4.00mL of 2.00х10-3M Fe(NO3)3 and 3.00mL of 5.00х10-3M NaSCN and...

A solution was prepared by mixing 4.00mL of 2.00х10-3M Fe(NO3)3 and 3.00mL of 5.00х10-3M NaSCN and diluting the mixture with water to a total of 10.00mL. Use your average value of Kc to calculate the equilibrium concentration of FeSCN2+ in the mixture.

Average value of Kc is 787.3

Expert Solution

Initial moles of [Fe(H₂O)₆]³⁺ = 0.014 L * 2*10^-3 mol/L = 2.8*10^-5 mol
Initial moles of NaSCN = 3*10^-3 L * 5*10^-3 mol/L = 15*10^-6 mol
Make an ICE table,

  K_c = [FeSCN]²⁺ / [Fe³⁺] [SCN^-] = X / [2*10^-5 - X] [15*10^-6 - X]
787.3 = X / [2*10^-5 - X] [15*10^-6 - X]
  787.3 = X / [ 3*10^-10 - 2*10^-5X - 15*10^-6X + X² ]
787.3 [X² - 3.5*10^-5X + 3*10^-10] = X
787.3X² - 0.027555X + 2.3619*10^-7 = X
787.3X² - 1.0275555X + 2.3619*10^-7 = 0
   Solve the above quadratic expression,
   X = 2.2989*10^-7 mol = FeSCN²⁺
Concentration of FeSCN²⁺ at equilibrium = 2.2989*10^-7 mol / (0.014+3*10^-3)L = 1.3523*10^-5 mol/L
[FeSCN]²⁺ = 1.3523*10^-5M at equilibrium

queen_honey_blossom answered 2 years ago

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