The ionization constant of water and its ionic product
Posted: Wed Dec 02, 2020 6:48 am
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 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