5.0 5.0 5.0 5.0 5.0 1.0 2.0 3.0 4.0 5.0 4.0 3.0 2.0 1.0 Part B: Determination of

Question

5.0 5.0 5.0 5.0 5.0 1.0 2.0 3.0 4.0 5.0 4.0 3.0 2.0 1.0 Part B: Determination of the equilibrium constant Label 6 large test tubes and prepare solutions as outined in Table 6.2. The Fe solution is dispensed at 5 mL, for the other solutions use a 10 mL graduated pipet Table 6.2: Solutions to determine equilbrium constant. Note: The concentrations Ee and Fe: and HSCN are different from Part A 0.0030 MFe in 0.0030 M HSCN 0.50 M HNO Tube o50 M HNO, (mL) in 0.50 M HNO, (mL)(mL 5.0 5.0 5.0 5.0 5.0 5.0 0.0 1.0 2.0 3.0 4.0 5.0 5.0 4.0 3.0 2.0 1.0 Rel 10

Explanation / Answer

Tube-6:   

Initial moles of Fe3+(aq) = MxV = 0.0030 mol/L x 5.0x10-3L = 1.5x10-5 mol

Initial moles of HSCN = MxV = 0.0030 mol/L x 1.0x10-3L = 3.0x10-6 mol

Total volume of the solution = 10 mL = 1.0 x10-2 L

Initial moles of HNO3 = initial moles of H+(aq) = 0.5 mol/L x 1.0 x10-2 L = 5.0x10-3 mol

Equilibrium concentration of FeSCN2+(aq)= 6.23x10-5M  

Hence equilibrium moles of FeSCN2+(aq) = 6.23x10-5 mol/L x 1.0 x10-2 L = 6.23x10-7 mol

The balanced chemical reaction in equilibrium is

Fe3+(aq) + HSCN ---------- > FeSCN2+(aq) + H+  

Hence [Fe3+(aq)]eq = (1.5x10-5 mol - 6.23x10-7 mol) / 1.0 x10-2 L = 1.438x10-3 M (answer)

[HSCN]eq =  (3.0x10-6 mol - 6.23x10-7 mol) / 1.0 x10-2 L = 2.377x10-4 M (answer)

[H+]eq = (5.0x10-3 mol + 6.23x10-7 mol) / 1.0 x10-2 L = 0.5 M (answer)

Hence Kc = (6.23x10-5 M x 0.5 M) / (1.438x10-3 M x 2.377x10-4 M) = 91.1 (answer)

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