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Handbook of Solvents - George Wypych - ChemTech - Ventech!

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630 Victor Cherginets<br />

controlled titrant loss affecting experimental results. Ditto referred to all results obtained by<br />

the reverse titration by MoO 3 and CrO 3 as the equivalence point.<br />

Acidic properties <strong>of</strong> P V oxocompounds in the chloride melt at 700 o C were investigated<br />

using a membrane oxygen electrode Ni,NiO|ZrO 2. 93 Polyphosphates <strong>of</strong> compositions with<br />

Na 2O-P 2O 5 ratio from 1.67 to 3 formed in the acidic region:<br />

− 2 n+<br />

2<br />

3−<br />

PO + PO = P O<br />

( ) − ( n+<br />

5)<br />

n 3n+ 1<br />

4 n+ 1 3n+ 5<br />

[10.4.43]<br />

The corresponding constants were estimated. Polyphosphate solubilities were determined<br />

by a cryoscopic method.<br />

The titration <strong>of</strong> V 2O 5 proceed in two stages: 94<br />

pK 9303 . . pK 8303 . .<br />

VO ⎯⎯⎯⎯⎯→VO⎯⎯⎯⎯⎯→VO 2 5<br />

=− ± − =− ± 4−<br />

3<br />

2 7<br />

[10.4.44]<br />

Some results <strong>of</strong> oxoacidity studies in molten KCl-NaCl were obtained in our<br />

works, 95-99 they are presented in Table 10.4.1.<br />

Table 10.4.1. The Lux acid-base equilibrium constants in molten KCl-NaCl and NaI (at<br />

the confidence level 0.95) [After references 95,96,98]<br />

Equilibrium<br />

m<br />

KCl-NaCl<br />

-pK<br />

N mol%<br />

- 2−<br />

4−<br />

2PO3 + O = P2O7 8.01±0.1 10.67 6.67<br />

− 2−<br />

3−<br />

PO3 + O = PO4<br />

5.93±0.1 7.26 5.26<br />

− 2−<br />

3− 2−<br />

PO3 + 2O = [ PO4 ⋅O]<br />

7.24±0.1 - -<br />

2−<br />

2−<br />

CrO3 + O = CrO4<br />

8.41±0.1 9.77 7.77<br />

2− 2−<br />

2−<br />

Cr2O7 + O = 2CrO4<br />

7.18±0.1 7.18 7.18<br />

2− 2−<br />

2− 2−<br />

2CrO4 + O = [2CrO4 ⋅O ]<br />

1.60±0.2 4.26 0.26<br />

2−<br />

2−<br />

MoO3 + O = MoO4<br />

8.32±0.2 9.65 7.65<br />

2−<br />

2− 2−<br />

MoO3 + 2O = [MoO4 ⋅O<br />

]<br />

9.71±0.3 11.37 7.37<br />

2−<br />

2−<br />

WO3 + O = WO4<br />

9.31±0.2 10.64 8.64<br />

2−<br />

2− 2−<br />

WO3 + 2O = [WO4 ⋅O<br />

]<br />

10.67±0.5 13.33 9.33<br />

2− 2−<br />

−<br />

BO 4 7 + O = 4BO2<br />

4.82±0.1 2.16 4.16<br />

− 2−<br />

3−<br />

BO2 + O = BO3<br />

2.37±0.2 3.17 1.70<br />

2−<br />

−<br />

VO 2 5 + O = 2VO3<br />

6.95±0.2 6.95 6.95<br />

2−<br />

4−<br />

VO 2 5 + 2O=<br />

VO 2 7<br />

12.23±0.1 14.89 10.89<br />

2−<br />

3−<br />

VO 2 5 + 3O=<br />

2VO4<br />

12.30±0.1 14.96 10.96<br />

2−<br />

3− 2−<br />

VO 2 5 + 5O= 2[VO4 ⋅O<br />

]<br />

13.88±0.5 19.20 11.20<br />

2GeO<br />

2−<br />

2−<br />

+ O = Ge O<br />

4.18±0.4 6.84 2.84<br />

2<br />

2 5

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