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Why does $\ce{H2Se}$ have a bigger $K_\text{a2}$ than both $\ce{H2S}$ and $\ce{H2Te}$?

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    $\begingroup$ What do you mean with "$Ka_2$"? $\endgroup$ – Philipp Mar 24 '15 at 19:40
  • $\begingroup$ @Philipp $K_{a2}$ refers to the second dissociation constant i.e. the equilibrium constant of $\ce {HA^{-}<=> H+ + A^{2-}}$ $\endgroup$ – Binary Geek Mar 24 '15 at 19:46
  • $\begingroup$ Would you mind to provide the data and the source? $\endgroup$ – Klaus-Dieter Warzecha Mar 24 '15 at 20:01
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The following data are from the Rubber Book (CRC Handbook of Chemistry and Physics), 89th edition, 2009, p 8-40.

\begin{array}{lll} \mathrm{acid} & \mathrm{pK_{a1}} & \mathrm{pK_{a2}} \\ \hline \ce{H2S} & 7.05 & 19\\ \ce{H2Se} & 3.89& 11\\ \ce{H2Te} & 2.6 & 11\\ \end{array}

Note that $\mathrm{pK_{a1}}$ of $\ce{H2Te}$ was determined at 18 °C, all other data at 25 °C

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