WBJEE · Chemistry · Ionic Equilibrium
At \(25^{\circ} \mathrm{C}\), the ionic product of water is \(10^{-14}\). The free energy change for the self-ionization of water in \(\mathrm{kCal}^{-1} \mathrm{~mol}^{-1}\) is close to
- A 20.5
- B 14
- C 19.1
- D 25.3
Answer & Solution
Correct Answer
(C) 19.1
Step-by-step Solution
Detailed explanation
\(\mathrm{H}_2 \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad \mathrm{K}=10^{-14}\)…
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