5.2. Titration in NaOH (C, V) ) HIO (C0, V0) system

In the system II, disproportionation of HBrO affected by NaOH (C) added according to the

Figure 2. (A) pH = pH(Φ) and (B) E = E(Φ) relationships plotted for the systems I and II, and the related speciation

102.5563 at Φ = 2.0; 102.2730/102.5740 at Φ = 2.5, i.e. 100.3010 = 2 = 2:1, corresponding to the stoichiometric ratio of products of this reaction. As results from Figure 2C, the disproportionation, at an excess of NaOH added, occurs mainly according to reaction 3HBrO + 3OH<sup>1</sup> =

<sup>1</sup> + 3H2O (stoichiometry 3:3 = 1:1), resulting from half reactions: HBrO + 2e<sup>1</sup> +

]/[BrO3 1

<sup>1</sup> + 5H+1, and 3H+1 + 3OH<sup>1</sup> = 3H3O. The

] ratio equals: 102.2553/

titrimetric mode is presented in Figure 2D [16]. The [Br<sup>1</sup>

H+1 = Br<sup>1</sup> + H2O, HBrO – 4e<sup>1</sup> + 2H2O = BrO3

diagrams for the systems: I (C) and II (D).

26 Redox - Principles and Advanced Applications

2Br<sup>1</sup> + BrO3

The curves plotted in Figure 3a–c are related to titration of V<sup>0</sup> = 10 mL of HIO (C<sup>0</sup> = 0.1 mol/L) with V mL of C = 0.1 mol/L NaOH. At the initial part of the titration, we have the reactions:

Figure 3. Plots of (a) E = E(Φ), (b) pH = pH(Φ), and (c) speciation curves for NaOH (C, V) ) HIO (C0, V0) system at V<sup>0</sup> = 10, C<sup>0</sup> = 0.01, C = 0.1.

$$\text{5HIO} + \text{OH}^{-1} = \text{2}(\text{I}\_{2(s)}\text{I}\_2) + \text{IO}\_3^{-1} + \text{3H}\_2\text{O} \tag{50}$$

In the following, at Φ ca. 0.20–0.22, a pronounced increase in [I �1 ] occurs, as a result of reaction

$$\text{2\ 3HIO} + \text{3OH}^{-1} = \text{IO}\_3^{-1} + \text{2I}^{-1} + \text{3H}\_2\text{O} \tag{51}$$

The increase in [I �1 ] is accompanied by an increase in [I<sup>3</sup> �1 ]

$$I \left( I\_{\mathbf{2}(s)}, I\_2 \right) \; + I^{-1} = I\_3^{-1} \tag{52}$$

This leads to the gradual disappearance of I2(s) (which is ultimately ended at Φ = 0.5347) and lowering of [I2] and [I<sup>3</sup> �1 ]; all them disproportionate

$$6\text{ }\text{\textbullet}(\text{I}\_{2(s)}\text{ }I\_2\text{ }I\_3^{-1}) + 6\text{OH}^{-1} = \text{IO}\_3^{-1} + \text{(5, 5, 8)}I^{-1} + 3\text{H}\_2\text{O} \tag{53}$$

At [I2(s)] > 0, we have [I2] = <sup>s</sup> = const; <sup>s</sup> = 1.33�10�<sup>3</sup> mol/L is the solubility of <sup>I</sup>2(s) in water, at 20�C. Finally, the disproportionation of HIO, affected by NaOH, can be expressed by the equation

$$\text{3HIO} + \text{3OH}^{-1} = \text{IO}\_3^{-1} + 2I^{-1} + \text{3H}\_2\text{O} \tag{54}$$

Note that the stoichiometry of the reaction (54) is 3 : 3 = 1 : 1, which corresponds to the jump on the curves presented in Figures 3a,b, occurring at Φ = 1. For Φ > 1, we have [I�<sup>1</sup> ]/[IO3 �1 ] = 2, i.e. the stoichiometry of the products of reaction (54) equals to 1 : 2. The jumps on the curves E = E(Φ) and pH = pH(Φ) (Figures 3a,b) occur at Φ ca. 0.2 (which corresponds to the stoichiometry 1:5 of the reaction (50)) and at Φ ca. 1 (which corresponds to the stoichiometry 3:3 = 1:1 of the reaction (54)); the maxima on the corresponding derivative curves in Figures 4a,b fit the stoichiometric ratios.
