Despite the structural differences with $H_{2} S O_{4}$ and selenic acid, telluric acid is a diprotic acid with $p k a_{1} = 7.68$ and $p k a_{2} = 11.29.$ Telluric acid can be prepared by hydrolysis of tellurium hexafluoride
$T e F_{6} + 6 H_{2} O \to T e {(O H)}_{6} + 6 H F$
$T e F_{6}$ can be prepared by the reaction between elements
$T e + 3 F_{2} \to T e F_{6}$
A cube of tellurium (density $6 g / c c$ ) measuring $1.2 c m$ on edge is allowed to react with $2.34 L$ of fluorine gas at 1 atm and $27 ° C$ . What is the pH of the solution considering telluric acid only (Neglect HF and considered as volume of solution is 1 L). Assume $100 %$ yield in all reactions.
$(A t W t : T e = 128, F = 19, O = 16, H = 1)$ $(\log 3.17 = 0.5)$

Question

Despite the structural differences with $H_{2} S O_{4}$ and selenic acid, telluric acid is a diprotic acid with $p k a_{1} = 7.68$ and $p k a_{2} = 11.29.$ Telluric acid can be prepared by hydrolysis of tellurium hexafluoride
$T e F_{6} + 6 H_{2} O \to T e {(O H)}_{6} + 6 H F$
$T e F_{6}$ can be prepared by the reaction between elements
$T e + 3 F_{2} \to T e F_{6}$
A cube of tellurium (density $6 g / c c$ ) measuring $1.2 c m$ on edge is allowed to react with $2.34 L$ of fluorine gas at 1 atm and $27 ° C$ . What is the pH of the solution considering telluric acid only (Neglect HF and considered as volume of solution is 1 L). Assume $100 %$ yield in all reactions.
$(A t W t : T e = 128, F = 19, O = 16, H = 1)$ $(\log 3.17 = 0.5)$

Accepted Answer

Number of moles of  F2=1×2.30.0821×300=0.095Mass of  Te=1.2×1.2×1.2×6=10.368Number of moles of  Te+3F2→TeF6Number of moles of  Te=10.368128=0.081  Limiting agent is  F23 moles of  F2 gives 1 moles of TeF6 0.095 moles of F2  gives  3.1×10−2moles of TeF63.1×10−2moles of TeF6 or Te(OH)6 pH=pka−logC2

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