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# What is the equilibrium constant for water?

Last updated date: 21st Jul 2024
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Hint : As we know that Standard electrode potential is the potential at standard conditions. The standard EMF of a reaction is obtained by taking the difference of standard potential of cathode and anode. Cathode is the right electrode and anode is the right electrode also water is amphoteric in nature.

When two-electrodes are dipped in solution containing their electrolyte solution, the more reactive electrodes undergo oxidation whereas the less reactive electrodes undergo reduction also there is a tendency of metal atoms to go into the solution. There are some substances like water are amphoteric in nature which means they can act as both acid and base so in the case of water let one molecule of water act as acid and other one molecule act as base so the acidic molecule will donate a proton and the base will accept the proton so the following equilibrium will exist: ${{H}_{2}}O(l)+{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)$
The dissociation constant for this reaction will be $K=\dfrac{[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}$
here ${{K}_{w}}=k[{{H}_{2}}O]$ Now this is how the constant is made.
Now here the concentration of water is being omitted from the denominator as water is a pure liquid and its concentration remains constant. So concentration of water will be incorporated within the equilibrium constant to give a new constant, ${{K}_{w}}$ which is also known as ionic product of water.
${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
Remember to write an expression for the equilibrium constant for the formation of the water gas. $K=\dfrac{[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}$ In the above equilibrium constant expression, the concentration of carbon solid does not appear. In an equilibrium constant expression, the concentration of solids and pure liquids does not appear.