Answer:
5x10⁻⁶ = [HTeH₄O₆⁺]
Explanation:
The first dissociation equilibrium of the telluric acid in water is:
H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺
Using H-H equation for telluric acid:
<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>
pKa of telluric acid is -logKa1
pKa = -log 2.0x10⁻⁸
pKa = 7.699
As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:
3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]
-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]
2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]
<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>
