SODIUM IODIDE NaI

Properties of sodium iodide NaI:

White, when stored in the light turn yellow due to oxidation. It melts without decomposition, on further heating decomposes. Readily soluble in water (no hydrolysis). Typical reductant.


Obtaining sodium iodide NaI:

2 Na + E2 = 2 NaE (normal temp., E = F, Cl; Вr, I).

2 NaOH (hot) + I2 + H2O2 =2 NaI + O2↑ + 2H2O

2 NaOH (cold) + I2 + H2S(gas) = 2 NaI + S↓ + 2H2O.

2 NaOH (diluted) + FeI2 = 2NaI + Fe(OH)2↓ (in the atm. N2)

2 NaOH (diluted) + 2AgNO3 = Ag2O ↓ + H2O + 2NaNO3.

NaIO3 + H2O + 2 Fe = NaI + 2 FeO(OH)↓ (boiling, on air).


Reactions with sodium iodide NaI:

2 NaI = 2Na + I2 (t > 1400° С).

NaI·2H2O = NaI + 2H2O (t > 68,9° С, vacuum).

NaI (diluted) + 4 H2O = [Na(H2O)4]+ + I- (pH = 7).

8NaI (solid) + 9 H2SO4 (conc.) = 4 I2↓ + H2S↑ + 4H2O + 8 NaHSO4 (30-50° С)

2NaI (solid) + 4 HNO3 (conc.) = I2↓ + 2NO2↑ + 2H2O + 2NaNO3 (boiling, impurity of HIO3).

6NaI + 2H2O + O2 → time → 4NaOH + 2Na[I(I)2] (normal temp.,on the light)

4NaI + 4 НСl (diluted.) + O2 = 2I2↓ + 4 NaCl + 2H2O (normal temp.,on the light).

2NaI (cold) + E2 = 2NaE + I2↓ (E = Cl, Br)

Nal (hot) + 3  H2O + 3Cl2 = NaIO3 + 6HCl.

NaI + 4 NH3 (liquid) = [Na(NH3)4]I↓ ↔ [Na(NH3)4]+ + I-

2 NaI + 3 H2SO4 (conc.) + MnO2 = 2NaHSO4 + I2↓ + 2H2O.

NaI + 4 NaOH + 4 NaNO3 = Na5IO6 + 4NaNO2 + 2H2O (300-330° C).

NaI + 3H2O → Electrolysis → 3H2↑(on cathode) + NaIO3 (on anode).