5. In a dilute nitric acid solution, Fe reacts with thiocyanate ion (SCN) to for

Question

5. In a dilute nitric acid solution, Fe reacts with thiocyanate ion (SCN) to form a dark-red complex: The equilibrium concentration of [Fe(H2O)NCSmay be determinedby how darkly colored the solution is (measured by a spectrometer). In one such experiment, 1.00 mL of 0.200 MFe(NOs)h was mixed with 1.00 mL of 1.00x 10 MKSCN and 8.00 mL of dilute HNO,. The color of the solution quantitatively indicated that the [Fe(HO)NCS concentration was 7.30 × 10-5 M. Calculate the formation constant for [Fe(H2O)sNCS 6. Which of the following hydrated cations are colorless: Fe? (ag), Zn (a), Cu (a) Cu'(ag), V (ag), Ca (ag), Co (ag), Se (a), Pb2 (aq)? Explain your choice. 7. Manganese forms three low-spin complex ions with the cyanide ion with the formulas oxidation number of Mn and the number of unpaired d electrons present. Oxidation number of Mn Number of unpaired d electrons Complex ion [Mn(CN)] Mn(CN1 [Mn(CN)6])-

Explanation / Answer

5. The reaction is given as

[Fe(H2O)6]3+ + SCN- ç===è H2O + [Fe(H2O)5NCS]2+; this is an equilibrium system and the equilibrium constant is given as

K = [Fe(H2O)5NCS]2+/[Fe(H2O)6]3+[SCN-] (The concentration of pure substances are ignored while writing equilibrium constants for dissociation reactions). This equilibrium constant is the formation constant.

To evaluate K, we will need to calculate the equilibrium concentrations of the ionic species in the solution. While calculating the concentrations, we must keep in mind the total volume of the solution.

Total volume of the solution is (1+1+8) mL = 10 mL. Let us calculate the molarity of the ionic species in the solution.

Moles of [Fe(H2O)6]3+ added = (0.001 L) (0.200 mol/L) = 0.0002 mole (we start with 1.0 ml of Fe(NO3)3).

Moles of KSCN added = (0.001 L)(1.0*10-3 mol/L) = 1.0*10-6 mole (we start with 1.0 ml of KSCN).

Note in aqueous solution, Fe3+ always forms the hydrated ion. We next calculate the molarity of the ionic species in the solution as

Molarity of [Fe(H2O)]3+ in the solution = (0.0002 mol/ 0.01 L) = 0.02 M (total volume of solution is 10 mL)

Molarity of KSCN = (1.0*10-6 mol/0.01 L) = 1.0*10-4 M

We set up the ICE chart as below:

[Fe(H2O)6]3+ + SCN- ç===è H2O + [Fe(H2O)5NCS]2+

Initial                  0.02         1.0*10-4                                       0

Change                -x                 -x                                              x

Equilibrium (0.02 –x) (1.0*10-4 – x)                                    x

Hence, K = (x)/(0.02 – x)(1.0*10-4 – x)

The concentration of [Fe(H2O)5NCS]2+ is 7.30*10-5 M.

Hence, K = (7.30*10-5 M)/[(0.02 – 7.30*10-5)M][(1.0*10-4 – 7.30*10-5)M]

Or, K = (7.30*10-5)/(0.019927)(0.000027) M-1

Or, K = 140.62 (ans)

6. The colour of transition metal ions in aqueous solution is due to the migration of electron(s) within the available d-orbitals of the metal. This is known as d-d transition and is responsible for faint to dark colouration of the aqueous solutions. Point to remember here is that there must be vacant d-orbitals available for the transitions. With this logic in mind, we would expect the hydrated ions of Ca2+ (at. no 20, no d electrons) and Zn2+ (at. no 30, completely filled 3d orbitals) to be colourless. All the other ions have one or more available d electrons and vacant orbitals for the transitions.

7. We must assume that the oxidation number of CN- is (-1). Let the oxidation number (O.N.) of Mn in [Mn(CN)6]5- be a.

Therefore, a + 6.(-1) = -5 (as the complex carries an overall negative charge of -5 and all charges must be balanced).

or, a – 6 = -5

or, a = +1

We follow the similar approach for [Mn(CN)6]4-. Let the O.N. of Mn be b.

b – 6 = -4

or, b = +2

Similarly, for [Mn(CN)6]3-, we have

c – 6 = -3 where c is the O.N. of Mn

or c = +3

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