Chemistry Help?

After completing Part A of this experiment, a student determines the slope of their calibration graph of absorbance versus [FeSCN2] to be 5140 M1. In Part B, a student combines 2.00 mL of a 0.00200 M solution of KSCN with 5.00 mL of 0.00200 M solution of Fe(NO,), and 3.00 mL of water. 

The following equilibrium is established.

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

The absorbance of the equilibrium solution is measured as 0.246. What is the equilibrium constant for this reaction?

1 Answer

  • 4 weeks ago

    For this reaction, Kc = [FeSCN2+]/[Fe3+][SCN-]

    First, you can use Beer's law to calculate the molarity and moles of FeSCN2+ in the solution:

    A = ebc

    0.246 = 5140 (1) (c)

    c = 4.79X10^-5 M = [FeSCN2+]

    Moles FeSCN = 4.79X10^-5 mol/L X 0.0100 L = 4.79X10^-7 moles FeSCN2+

    Next, calculate the original moles of Fe3+ and SCN- in the solution, and then their final concentrations:

    Original moles Fe3+ = 0.00200 L X 0.00200 mol/L = 4.00X10^-6 moles

    Moles Fe3+ at equilibrium = 4.00X10^-6 - 4.79X10^-7 = 3.52X10^-6 mol

    [Fe3+] at equilibrium = 3.52X10^-6 mol / 0.0100 L = 3.52X10^-4 M Fe3+

    Original moles SCN- = 0.00500 L X 0.00200 mol/L = 1.00X10^-5 mol

    Moles SCN- at equilibrium = 1.00X10^-5 - 4.79X10^-7 = 9.52X10^-6 mol

    [SCN-] at equilibrium = 9.52X10^-6 mol / 0.0100 L = 9.52X10^-4 M SCN-

    Kc = 4.79X10^-5 / (3.52X10^-4) (9.52X10^-4) = 143

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