The following equilibrium exists.\(H_2O(l)_{\text{acid}} +H_2O(l)_{\text{base}}\rightleftharpoons H_3O^+(aq)_{\text{Conjugate acid}} + OH^-(aq)_{\text{Conjugate base}}\)
The equilibrium constant\((K)=\frac{[H_3O^+][OH^-]}{[H_2O]}=[H^+][OH^-]=K_w=?\)
Now, how is \( (K_w)= {[H_3O^+][OH^-]}\)= (1.00e-7)² M =1.00e-14 M²
The ionization constant of water and its ionic product
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- Dhamnekar Winod
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The ionization constant of water and its ionic product
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- ChenBeier
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Re: The ionization constant of water and its ionic product
Its because the dissociation is like this. In neutral water 10^-7 H+ and also the same amount of OH- in solution.