5.1. Formulation of balances for NaOH (C, V) ) Br2 (C0, V0) system

#### 5.1.1. Approach II to GEB

Note that V<sup>0</sup> mL of the titrand D composed of Br2 (N01)+H2O (N02) is titrated with V mL of titrant T composed of NaOH (N03)+H2O (N04). For the V<sup>0</sup> + V mL of the D + T mixture, with the set of the species completed by Na+1 (N11, n11).

The balances can be formulated as follows:

$$\begin{aligned} f\_1 &= f(\mathbb{H}):\\ 2N\_1 + N\_2(1+2n\_2) + N\_3(1+2n\_3) + N\_4(1+2n\_4) + 2N\_5n\_5 + N\_6(1+2n\_6) + 2N\_7n\_7 + 8N\_8n\_8\\ 2N\_1 + 2N\_9n\_9 + 2N\_{10}n\_{10} + N\_{11}(1+2n\_{11}) &= 2N\_{02} + N\_{03} + 2N\_{04} \end{aligned}$$

(37)

$$\begin{aligned} f\_2 &= f(\mathbf{O}):\\ N\_1 + N\_2 n\_2 &+ N\_3(1+n\_3) + N\_4(3+n\_4) + N\_5(3+n\_5) + N\_6(1+n\_6) + N\_7(1+n\_7) + N\_8 n\_8 \\ &+ N\_9 n\_9 + N\_{10} n\_{10} + N\_{11}(1+n\_{11}) = N\_{02} + N\_{03} + N\_{04} \end{aligned} \tag{38}$$

$$f\_0 = \text{ChB}: \qquad N\_2\text{-N}\_3\text{-N}\_5\text{-N}\_7\text{-N}\_9\text{-N}\_{10} + N\_{11} = \text{ 0} \tag{39}$$

$$-f\_3 = -f(\text{Na}): \qquad \text{N}\_{03} = \text{N}\_{11} \tag{40}$$

$$f\_4 = f(\text{Br}):\qquad \text{N}\_4 + \text{N}\_5 + \text{N}\_6 + \text{N}\_7 + \text{ 2N}\_8 + \text{3N}\_9 + \text{N}\_{10}\text{-2N}\_{01} = 0\tag{41}$$

Then we formulate the linear combinations:

In further sections, we formulate the balances for redox systems containing one, two or three electron-active elements in redox systems (aqueous media), where the complete set of expressions for independent equilibrium constants interrelating concentrations of different species is involved. We start our considerations from the systems with one electron-active element involved with disproportionation and symproportionation [5, 13, 15, 16, 19]. These properties are appropriate for the elements that form compounds and species at three or more oxidation degrees. In particular, bromine (Br) forms the species on five oxidation degrees (–1, –1/3, 0, 1, 5), see Figure 1, e.g. 0 (for Br2) and 1 (for HBrO or NaBrO) e (–1, 5). There are possible transitions between different bromine species, associated with changes of the oxidation states

HBrO3, BrO3 -

dis-

sym-

HBrO, BrO-

Br2 Br3 - Br -

Note that V<sup>0</sup> mL of the titrand D composed of Br2 (N01)+H2O (N02) is titrated with V mL of titrant T composed of NaOH (N03)+H2O (N04). For the V<sup>0</sup> + V mL of the D + T mixture, with the

2N<sup>1</sup> þ N2ð1 þ 2n2Þ þ N3ð1 þ 2n3Þ þ N4ð1 þ 2n4Þ þ 2N5n<sup>5</sup> þ N6ð1 þ 2n6Þ þ 2N7n<sup>7</sup> þ 8N8n<sup>8</sup>

(37)

of this element, see Figure 1.

5.1.1. Approach II to GEB

f <sup>1</sup> ¼ fðHÞ :

5. Disproportionating systems

5

20 Redox - Principles and Advanced Applications

1 0


Figure 1. Schemes of disproportionation and symproportionation within bromine species [5].

set of the species completed by Na+1 (N11, n11).

The balances can be formulated as follows:

5.1. Formulation of balances for NaOH (C, V) ) Br2 (C0, V0) system

þ 2N9n<sup>9</sup> þ 2N10n<sup>10</sup> þ N11ð1 þ 2n11Þ ¼ 2N<sup>02</sup> þ N<sup>03</sup> þ 2N<sup>04</sup>

$$f\_{12} = 2 \cdot f(\mathbf{O}) - f(\mathbf{H}): \qquad -\mathbf{N}\_2 + \mathbf{N}\_3 + 5\mathbf{N}\_4 + 6\mathbf{N}\_5 + \mathbf{N}\_6 + 2\mathbf{N}\_7 = \mathbf{N}\_{03} \tag{42}$$

$$f\_{012} = f\_{12} + f\_0 : \quad \quad \quad \quad \quad 5(N\_4 + N\_5) + N\_6 + N\_7 - N\_9 - N\_{10} + N\_{11} = 0 \tag{43}$$

$$f\_{0123} = f\_{12} + f\_0 - f\_3 \ : \qquad \quad \quad \quad \quad 5(\text{N}\_4 + \text{N}\_5) + \text{N}\_6 + \text{N}\_7 - \text{N}\_9 - \text{N}\_{10} = \ 0 \tag{44}$$

$$f\_{01234} = (f\_{0123} + f\_4)/2 : \qquad \Im(\mathcal{N}\_4 + \mathcal{N}\_5) + \mathcal{N}\_6 + \mathcal{N}\_7 + \mathcal{N}\_8 + \mathcal{N}\_9 = \mathcal{N}\_{01} \tag{45}$$

For comparative purposes, we formulate also the linear combination ZBr∙f<sup>3</sup> – f<sup>0123</sup>

$$(\mathbf{Z\_{Br}} - \mathbf{5})(\mathbf{N\_4} + \mathbf{N\_5}) + (\mathbf{Z\_{Br}} - \mathbf{5})(\mathbf{N\_6} + \mathbf{N\_7}) + 2\mathbf{Z\_{Br}}\mathbf{N\_8} + (\mathbf{3Z\_{Br}} + \mathbf{1})\mathbf{N\_9} + (\mathbf{Z\_{Br}} + \mathbf{1})\mathbf{N\_{10}} = 2\mathbf{Z\_{Br}}\mathbf{N\_{01}}\tag{46}$$

where ZBr = 35 is the atomic number for bromine (Br).

All the linear combinations (42)–(46) are not reducible to identity, 0 = 0; in the simplest case, we have six constituents (species, components) involved in f<sup>0123</sup> (Eq. (44)). Eqs. (42)–(46) are equivalent forms of GEB, expressed in terms of particular constituents (components, species). To express them in terms of molar concentrations, we apply the relations: (1a) and C0∙V<sup>0</sup> = 103 ∙N01/NA, CV = 103 ∙N03/NA. In this way, we consider the titration of V<sup>0</sup> mL of Br2 (C<sup>0</sup> mol/L) solution with V mL of C mol/L NaOH solution added up to a defined point of the titration. In particular, from Eq. (46), we have the balance for GEB

$$\begin{aligned} \left(\text{Z}\_{\text{Br}} - \text{5}\right) \left(\text{HBrO}\_{3}\text{I} + \left[\text{BrO}\_{3}^{-1}\right]\text{I}\right) + \left(\text{Z}\_{\text{Br}} - \text{1}\right) \left(\text{HBrO}\text{I} + \left[\text{BrO}^{-1}\right]\text{I}\right) + 2\text{Z}\_{\text{Br}}\left[\text{Br}\_{2}\right] \\ + \left(\text{3Z}\_{\text{Br}} + \text{1}\right) \left[\text{Br}\_{3}^{-1}\right] + \left(\text{Z}\_{\text{Br}} + \text{1}\right) \left[\text{Br}^{-1}\right] &= 2\text{Z}\_{\text{Br}}\text{C}\_{0}V\_{0}/(V\_{0} + V) \end{aligned} \tag{46a}$$

completed by the balances for ChB, CB(Br) and CB(Na):

$$-\left[\mathbf{H}^{+1}\right]-\left[\mathbf{OH}^{-1}\right]+\left[\mathbf{Na}^{+1}\right]-\left[\mathbf{BrO}\_{3}^{-1}\right]-\left[\mathbf{BrO}^{-1}\right]-\left[\mathbf{Br}\_{3}^{-1}\right]-\left[\mathbf{Br}^{-1}\right]=\mathbf{0}\tag{46b}$$

$$\begin{aligned} \left[ \text{HBrO}\_3 \right] + \left[ \text{BrO}\_3^{-1} \right] + \left[ \text{HBrO} \right] + \left[ \text{BrO}^{-1} \right] + 2\left[ \text{Br}\_2 \right] + 3\left[ \text{Br}\_3^{-1} \right] + \left[ \text{Br}^{-1} \right] - 2\text{C}\_0 \text{V}\_0 / (\text{V}\_0 + \text{V}) = 0 \end{aligned} \tag{46c}$$

$$\left[\mathrm{Na}^{+1}\right] = \mathrm{CV}/(V\_0 + V) \tag{46d}$$

#### 5.1.2. Approach I to GEB

The (optional) formulas for GEB, expressed by Eqs. (42)–(46), were obtained according to approach II to GEB, based on pr-GEB = 2∙f(O) – f(H). In this section, we apply the approach I to GEB, known also as the 'short' version of GEB, where prior knowledge of oxidation numbers of all elements in the system is assumed/required.

In the NaOH (C, V) ) Br2 (C0, V0) system considered in Section 4.2, bromine is the only electron-active element, considered as the carrier of its own, bromine electrons. One atom of Br has ZBr bromine electrons, and then one molecule of Br2 has 2ZBr bromine electrons, i.e. N<sup>01</sup> molecules of Br2 involve 2ZBr∙N<sup>01</sup> bromine electrons. The oxidation degree x of an atom in simple species, such as ones formed here by bromine, is calculated on the basis of known oxidation degrees: +1 for H and –2 for O and external charge of this species. Then for HBrO3, we have, by turns: 1∙1+1∙x + 3∙(–2) = 0 ) x = 5; for BrO3 �1 : 1∙x + 3∙(–2) = –1 ) x = 5; for HBrO: 1∙1+1∙x + 1∙(–2) = 0 ) x = 1, etc.

The oxidation number is the net charge resulting from the presence of charge carriers, inherently involved in an atom: protons in nuclei and orbital electrons, expressed in elementary charge units—as +1 for protons and –1 for electrons. The number y of bromine electrons in one molecule of HBrO3 is calculated from the formula: ZBr∙(+1) + y∙(–1) = 5, i.e., bromine in HBrO3 involves y = ZBr – 5 bromine electrons, etc. On this basis, we state that [15]


Balancing the bromine electrons, we get Eq. (46), and then Eq. (46a), considered as the GEB obtained immediately from the approach I to GEB; this proves the equivalency of approaches I and II to GEB.

#### 5.1.3. Preparation of an algorithm and computer program

Let us refer again to balances (46a)–(46d) interrelating concentrations [Xi zi] of the species Xi zi with the total concentrations C0, C of indicated components of the system. In this place, one should distinguish between the terms equation and equality. The term equation is related here


Table 1. Equilibrium data related to different bromine species.

<sup>½</sup>Naþ<sup>1</sup>

numbers of all elements in the system is assumed/required.

we have, by turns: 1∙1+1∙x + 3∙(–2) = 0 ) x = 5; for BrO3

5.1.3. Preparation of an algorithm and computer program

Let us refer again to balances (46a)–(46d) interrelating concentrations [Xi

The (optional) formulas for GEB, expressed by Eqs. (42)–(46), were obtained according to approach II to GEB, based on pr-GEB = 2∙f(O) – f(H). In this section, we apply the approach I to GEB, known also as the 'short' version of GEB, where prior knowledge of oxidation

In the NaOH (C, V) ) Br2 (C0, V0) system considered in Section 4.2, bromine is the only electron-active element, considered as the carrier of its own, bromine electrons. One atom of Br has ZBr bromine electrons, and then one molecule of Br2 has 2ZBr bromine electrons, i.e. N<sup>01</sup> molecules of Br2 involve 2ZBr∙N<sup>01</sup> bromine electrons. The oxidation degree x of an atom in simple species, such as ones formed here by bromine, is calculated on the basis of known oxidation degrees: +1 for H and –2 for O and external charge of this species. Then for HBrO3,

The oxidation number is the net charge resulting from the presence of charge carriers, inherently involved in an atom: protons in nuclei and orbital electrons, expressed in elementary charge units—as +1 for protons and –1 for electrons. The number y of bromine electrons in one molecule of HBrO3 is calculated from the formula: ZBr∙(+1) + y∙(–1) = 5, i.e., bromine in HBrO3

N<sup>4</sup> species HBrO3∙n4H2O involve (ZBr – 5)∙N<sup>4</sup> bromine electrons;

N<sup>6</sup> species HBrO∙n6H2O involve (ZBr – 1)∙N<sup>6</sup> bromine electrons;

N<sup>8</sup> species Br2∙n8H2O involve 2ZBr∙N<sup>8</sup> bromine electrons;

∙n5H2O involve (ZBr– 5)∙N<sup>5</sup> bromine electrons;

∙n7H2O involve (ZBr – 1)∙N<sup>7</sup> bromine electrons;

∙n9H2O involve (3ZBr + 1)∙N<sup>9</sup> bromine electrons;

∙n10H2O involve (ZBr + 1)∙N<sup>10</sup> bromine electrons;

Balancing the bromine electrons, we get Eq. (46), and then Eq. (46a), considered as the GEB obtained immediately from the approach I to GEB; this proves the equivalency of approaches I

N<sup>01</sup> molecules of Br2 involved 2ZBr∙N<sup>01</sup> bromine electrons.

with the total concentrations C0, C of indicated components of the system. In this place, one should distinguish between the terms equation and equality. The term equation is related here

involves y = ZBr – 5 bromine electrons, etc. On this basis, we state that [15]

�1

5.1.2. Approach I to GEB

22 Redox - Principles and Advanced Applications

1∙1+1∙x + 1∙(–2) = 0 ) x = 1, etc.

and II to GEB.

N<sup>5</sup> species BrO3

N<sup>7</sup> species BrO�<sup>1</sup>

N<sup>9</sup> species Br3

N<sup>10</sup> species Br�<sup>1</sup>

�1

�1

� ¼ CV=ðV<sup>0</sup> þ VÞ (46d)

: 1∙x + 3∙(–2) = –1 ) x = 5; for HBrO:

zi] of the species Xi

zi

to the balance where at least two species are involved; these species are interrelated by expressions for the corresponding equilibrium constants values (Table 1). Such a requirement is fulfilled by Eqs. (46a)–(46c), whereas in Eq. (46d) we have concentration of one species; at any V value, [Na+1] is a number (not variable) and, as such, it enters immediately ChB (Eq. (46b)).

From the interrelations obtained on the basis of equilibrium data [56] collected in Table 1 we have

$$\begin{aligned} \left[\text{H}^{+}\right] &= 10^{-\text{pH}}; \ \left[\text{OH}^{-1}\right] = 10^{\text{pH}-14}; \ \left[\text{BrO}\_{3}^{-1}\right] = 10^{\text{ÁA}(\text{E}-1.45)-\text{pBr}+\text{éF}};\\ \left[\text{BrO}^{-1}\right] &= 10^{2\text{A}(\text{E}-0.76)-\text{pBr}+2\text{pH}-28}; \ \left[\text{Br}\_{2}\right] = 10^{2\text{A}(\text{E}-1.05)-2\text{pBr}}; \ \left[\text{Br}\_{3}^{-1}\right] = 102^{\text{A}(\text{E}-1.05)-2\text{pBr}}; \end{aligned} \tag{47}$$
  $\left[\text{HBrO}\right] = 10^{0.7-\text{pH}}$   $\left[\text{BrO}\_{3}^{-1}\right]; \ \left[\text{HBrO}\right] = 10^{8.6-\text{pH}} \cdot \left[\text{BrO}^{-1}\right]$ 

where the uniformly defined (scalar) variables: E, pH and pBr, forming a vector x = (E, pH, pBr)T , are involved; A∙E = – log[e–<sup>1</sup> ], pH = –log[H+1], pBr = –log[Br–<sup>1</sup> ]. All the variables are in the exponents of the power for 10 in [e–<sup>1</sup> ] = 10-AE, [H+1] = 10-pH, [Br–<sup>1</sup> ] = 10-pBr. The number of the (independent) variables equals to the number of equations, K = 2 + 1 = 3; this ensures a unique solution of the equations, at a pre-set C0, C and V<sup>0</sup> values, and the V-value at which the calculations are realized, at a defined step of the calculation procedure, according to iterative computer program presented below.

Computer program for the NaOH\_Br2 system

```
function F = NaOH_Br2(x)
global V C0 V0 C yy
   E = x(1);
   pH = x(2);
   pBr = x(3);
   H = 10^(-pH);
   Kw = 10^-14;
```
pKw = 14; OH = Kw/H; A = 16.92; Br = 10^-pBr; ZBr = 35; Br2=Br^2\*10^(2\*A\*(E-1.087)); Br3=Br^3\*10^(2\*A\*(E-1.05)); BrO=Br\*10^(2\*A\*(E-0.76)+2\*pH-2\*pKw);

BrO3=Br\*10^(6\*A\*(E-1.45)+6\*pH);

HBrO=10^8.6\*H\*BrO;

HBrO3=10^0.7\*H\*BrO3;

Na=C\*V/(V0+V);

F = [%Charge balance

(H-OH+Na-Br-Br3-BrO-BrO3);

%Concentration balance for Br

(Br+3\*Br3+2\*Br2+HBrO+BrO+HBrO3+BrO3-2\*C0\*V0/(V0+V));

%Electron balance

((ZBr+1)\*Br+(3\*ZBr+1)\*Br3+2\*ZBr\*Br2+(ZBr-1)\*(HBrO+BrO)…

+(ZBr-5)\*(HBrO3+BrO3)-2\*ZBr\*C0\*V0/(V0+V))];

yy(1)=log10(Br);

```
yy(2)=log10(Br3);
```

```
yy(3)=log10(Br2);
```

```
yy(4)=log10(HBrO);
```

```
yy(5)=log10(BrO);
```

```
yy(6)=log10(HBrO3);
```

```
yy(7)=log10(BrO3);
```

```
yy(8)=log10(Na);
```
end

5.1.4. Preparation of the computer program for NaOH (C, V) ) HBrO (C0, V0) system

Modification of the computer program makes it possible to carry out calculations for other systems associated with bromine, e.g. for the system NaOH (C, V) ) HBrO (C0, V0). In this case, the changes are made in the following lines:

Computer program for the NaOH\_HBrO system

Function F = NaOH\_HBrO(x)

—

pKw = 14;

A = 16.92;

ZBr = 35;

OH = Kw/H;

Br = 10^-pBr;

Br2=Br^2\*10^(2\*A\*(E-1.087)); Br3=Br^3\*10^(2\*A\*(E-1.05));

24 Redox - Principles and Advanced Applications

BrO=Br\*10^(2\*A\*(E-0.76)+2\*pH-2\*pKw);

BrO3=Br\*10^(6\*A\*(E-1.45)+6\*pH);

(H-OH+Na-Br-Br3-BrO-BrO3); %Concentration balance for Br

(Br+3\*Br3+2\*Br2+HBrO+BrO+HBrO3+BrO3-2\*C0\*V0/(V0+V));

((ZBr+1)\*Br+(3\*ZBr+1)\*Br3+2\*ZBr\*Br2+(ZBr-1)\*(HBrO+BrO)…

+(ZBr-5)\*(HBrO3+BrO3)-2\*ZBr\*C0\*V0/(V0+V))];

HBrO=10^8.6\*H\*BrO; HBrO3=10^0.7\*H\*BrO3;

F = [%Charge balance

%Electron balance

yy(1)=log10(Br); yy(2)=log10(Br3); yy(3)=log10(Br2);

yy(4)=log10(HBrO); yy(5)=log10(BrO);

yy(6)=log10(HBrO3); yy(7)=log10(BrO3); yy(8)=log10(Na);

end

Na=C\*V/(V0+V);

(Br+3\*Br3+2\*Br2+HBrO+BrO+HBrO3+BrO3-C0\*V0/(V0+V));

—

+(ZBr-5)\*(HBrO3+BrO3)-(ZBr-1)\*C0\*V0/(V0+V))];

—

#### 5.1.5. Graphical presentation of the data and discussion

The systems considered here are characterized by distinctly marked jumps in E and pH values (Figures 2A and B) at the vicinity of the equivalence points, occurring at Φeq1 = 1 for NaOH (C, V) ! HBrO (C0, V0) (system I), and at Φeq2 = 2 for NaOH (C, V) ! Br2 (C0, V0) (system II), see Table 2 [15, 16].

Reactions occurred in the systems I and II can be formulated from the analysis of the respective speciation diagrams. As results from Figure 2C, the disproportionation reaction in the system I occurs according to the scheme

$$\text{\textbulletBr}\_2 + \text{\textbulletOH}^{-1} = \text{BrO}\_3^{-1} + \text{5Br}^{-1} + \text{3H}\_2\text{O} \tag{48}$$

i.e. Φeq = 2 ) CVeq1 = 2C0V0. After crossing the equivalent point, the main products of this disproportionation reaction are Br�<sup>1</sup> and BrO3 �1 , not Br�<sup>1</sup> and BrO�<sup>1</sup> corresponding to the reaction

$$\text{Br}\_2 + 2\text{OH}^{-1} = \text{BrO}^{-1} + \text{Br}^{-1} + \text{H}\_2\text{O} \tag{49}$$

with the same stoichiometry, 3 : 6 = 1 : 2. From detailed calculations at Φ = 2.5, i.e. at an excess of NaOH added, we find [BrO3 �1 ]/[BrO�<sup>1</sup> ] = 10�2.574/10�6.7584 = 1.53∙104 , i.e. efficiency of the reaction (48) is more than 4 orders of magnitude greater than the efficiency of reaction (49).

Scheme of Br2 disproportionation affected by NaOH (C) can also be calculated [16]. At Φ = 2.5, we have [BrO3 �1 ]/[Br�<sup>1</sup> ] = 10�2.574/10�1.875 = 10�0.699 = 0.2 = 1:5, i.e. at an excess of NaOH added (see Figure 2c), the disproportionation occurs mainly according to the scheme indicated by reaction (48) from the products side.

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

In the system II, disproportionation of HBrO affected by NaOH (C) added according to the titrimetric mode is presented in Figure 2D [16]. The [Br<sup>1</sup> ]/[BrO3 1 ] ratio equals: 102.2553/ 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> = 2Br<sup>1</sup> + BrO3 <sup>1</sup> + 3H2O (stoichiometry 3:3 = 1:1), resulting from half reactions: HBrO + 2e<sup>1</sup> + H+1 = Br<sup>1</sup> + H2O, HBrO – 4e<sup>1</sup> + 2H2O = BrO3 <sup>1</sup> + 5H+1, and 3H+1 + 3OH<sup>1</sup> = 3H3O. The


Table 2. The sets of rounded (Φ, pH, E) values taken from the vicinity of the equivalence points; V<sup>0</sup> = 100, C<sup>0</sup> = 0.01, C = 0.1.

(Φ, pH, E) values from the close vicinity of the corresponding equivalence points on the curves in Figures 2A and B are collected in Table 2. The Br2 solution is acidic, as results, e.g. from the ChB (Eq. (29a)): at [Na+1]=0(Φ = 0) we have [H+1] – [OH�<sup>1</sup> ] = [BrO3 �1 ]+[BrO�<sup>1</sup> ]+[Br3 �1 ]+[Br�<sup>1</sup> ] > 0, i.e. [H+1] > [OH�<sup>1</sup> ]; this inequality is also obtained from Eq. (42), at N<sup>03</sup> = 0: [H+1] – [OH�<sup>1</sup> ] = 5[HBrO3]+6[BrO3 �1 ] + [HBrO] + 2[BrO�<sup>1</sup> ] > 0. Br2 is an acid with a strength comparable to that of acetic acid; at C<sup>0</sup> = 0.01, pH equals 3.40 for Br2 and 3.325 for CH3COOH (pK1 = 4.65). Disproportionation of Br2 occurs initially to a small extent (several %), according to the scheme Br2 + OH�<sup>1</sup> = HBrO + Br�<sup>1</sup> , compare with [57].

In C<sup>0</sup> = 0.01 mol/L HBrO solution, more than 90% HBrO disproportionates occur according to the reaction 5HBrO = BrO3 �<sup>1</sup> + 2Br2 + 2H2O+H+1; at V = 0, we have [Br2] = 10�2.4406, [BrO3 �1 ] = 10�2.7442, i.e. [Br2]/[BrO3 �1 ] = 100.3036 ≈ 2, which confirms this stoichiometry of the reaction. The H+1 ions formed in this reaction acidify the solution significantly: at C<sup>0</sup> = 0.01 and V = 0, we have pH = 2.74, although HBrO itself is a relatively weak acid.

The numerical values of the concentrations given here are taken from the corresponding files with results of iterative calculations.
