## The ionization constant of water and its ionic product

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Dhamnekar Winod
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Location: Mumbai[Bombay],Maharashtra State,India

### The ionization constant of water and its ionic product

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
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
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Location: Berlin, Germany

### 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.