Table 4 Composition of equilibrium mixtures E1-E4 volume of 2.00 × 10-3M Fe(NOs)

Question

Table 4 Composition of equilibrium mixtures E1-E4 volume of 2.00 × 10-3M Fe(NOs)s in 10x 10 M HNO tolume of 2.00 × 10-3M NaSCN in 1.0x 10-'M HNO, solution, mL total solution volume, mL 100.00 100.00 100.00 100.00 flask number solution, mL E1 (blank) E2 ?? E4 50.00 50.00 50.00 50.00 0.00 10.00 30.00 50.00 Table 5 Analyzing the data flask number ?? E2 E4 initial Fe ion concentration, mol L-1 initial SCN" ion concentration, mol L- %T of equilibrium mixtures 72.9 43.2 32.5 absorbance of equilibrium mixture final equilibrium FeNCS+ion concentration, mol L-1 final equilibrium Fe ion concentration, mol L-1 final equilibrium SCN ion concentration, mol L- equilibrium constant, Ke mean Keq

Explanation / Answer

Calculations for E2:

Initial Fe3+ ion concentration = (2*10-3 mmol/mL * 50 mL)/100 mL = 10-3 mmol/mL = 10-3 mol/L

Initial SCN- ion concentration = (2*10-3 mmol/mL * 10 mL)/100 mL = 2*10-4 mol/L

%T = 72.9

i.e. T = 72.9/100 = 0.729

The absorbance of the solution (A) = -LogT = -Log(0.729) = 0.137

The concentration of SCN- ion consumed = 2*10-4*0.729 = 1.458*10-4 mol/L

The final equilibrium SCN- ion concentration = [SCN-]e = 2*10-4 - 1.458*10-4 = 5.42*10-5 mol/L

The final equilibrium Fe3+ ion concentration = [Fe3+]e = 10-3 - 1.458*10-4 = 8.542*10-4 mol/L

The final equilibrium FeNCS2+ concentration = [FeNCS2+]e = 1.458*10-4 mol/L

Now, the equilibrium constant, Keq = [FeNCS2+]e / [Fe3+]e*[SCN-]e

= 1.458*10-4/(8.542*10-4*5.42*10-5)

= 3.149*103

Similarly you can calculate the Keq values for E3 and E4.

The mean equilibrium constant = {(Keq)E2 + (Keq)E3 + (Keq)E4}/3

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