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Using the Thermodynamic Data Provided Below, Calculate Ka for HCN(aq) ∘\circ

Question 41

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Using the thermodynamic data provided below, calculate Ka for HCN(aq) at 25 ∘\circ . ΔHf∘( kJ/mol) S∘(J/K⋅ mol) H+(aq) 00CN−(aq) 151.0117.99HCN(aq) 105.4128.9\begin{array}{|l|c|c|}\hline & \Delta \mathrm{H}^\circ_{\mathrm{f}}(\mathrm{~kJ} / \mathrm{mol}) & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{~mol}) \\\hline \mathrm{H}^{+}(a q) & 0 & 0 \\\hline \mathrm{CN}^{-}(a q) & 151.0 & 117.99 \\\hline \mathrm{HCN}(a q) & 105.4 & 128.9 \\\hline\end{array}


A) 2.4 * 10-9
B) 2.6 * 10-9
C) 2.8 * 10-9
D) 3.0 * 10-9
E) None of the above

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