Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so that the pH is 3.00. What is the concentration of HTeH4O6

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:

pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]

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]

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explanation:

i think option no d is the correct answer