The Ionization of Water [closed]








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When we put $\ce{HCl}$ in a container of water and let's say that the concentration we put is $10^6$ molar. Let me understand this calculation: 


$[\ce{OH-}]=x$  

$[\ce{H+}]=10^6+x$  

$x(10^6+x)=10^{-14}$ 


The $[\ce{OH}]$ is $x$, but what about the $[\ce{H+}]=10^6+x$ why not $10^6-x$? Because the quantity of the $[\ce{H+}]$ does not increases after we add the $\ce{HCl}$ but decreases. So why $+x$ and not $-x$?




Consider the reaction,
$$\ce{HCl -> H+ + Cl-}$$


Assuming the initial $\ce{[HCl]}$ is $\pu{C M}$, after reaction takes place, we have $\ce{[H+]}$ as $\pu{C M}$.


Also its an aqueous solution. Hence we have to consider the following equilibrium:
$$\ce{ H2O <=> H+ + OH- }$$
Let initial $\ce{[H2O]}$ be $\ce{a M}$. At equilibrium, we have $\ce{[H+]} = \ce{x M}$. Since we have more $\ce{H+}$ from the ionization of water, therefore $\ce{[H+]\_{net}}$ = $\ce{C +x M}$ and not $\ce{C-x M}$


**Note**: If $C \geq \mathrm{10^{-6}~M}$, $\ce{[H+]}$ coming from $\ce{H2O}$, ie. $x$, can be neglected w.r.t $C$ unless and until you’re a rigid perfectionist.


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Your whole question is so muddled that it is difficult to know where to begin. 



> 
> When we put HCl in a container of water and let's say that the concentration we put is $10^6$ molar.
> 
> 
> 


$10^6$ molar HCl is totally absurd. The most concentrated HCl solutions will be around 12 molar. HCl is a gas at room temperature and pressure. To solidify HCl you must cool the gas to −114.22 °C. HCl has a molecular mass of 36.46 g/mol, thus neglecting any water, $10^6$ molar HCl would have $10^6$ moles of HCl in a liter volume, or a mass of 36460 kilograms per liter. Gold is only 19.30 kilograms per liter. 


The other thing is that HCl ionizes in water according to the reaction


$\ce{HCl + H2O -> H3O^+ + Cl^-}$


Since pure water is only about 56 molar, there isn't nearly enough water to ionize $10^6$ moles of HCl. 


**Let's talk about something more reasonable, 1.0 molar HCl.**


We can assume that all the HCl ionizes then the $\ce{[H^+]= 1.0 molar}$.


Since 


$\ce{[H^+][OH^-] = } 10^{-14}$


$\ce{[OH^-] = } \dfrac{10^{-14}}{\ce{[H^+]}} = \dfrac{10^{-14}}{1.0} = 1.0\times10^{-14}$


Now there is an assumption here about significant figures. So when we note that the HCl is 1.0 molar, we mean +/- 0.05 moles, or two significant figures. Very, very few chemistry experiments have results good to 4 significant figures. 


In reality the starting pure water had $1.0\times10^{-7}$ molar $\ce{[H^+]}$ and $1.0\times10^{-7}$ molar $\ce{[OH^-]}$. Since the final solution had $1.0\times10^{-14}$ molar $\ce{[OH^-]}$, then the added HCl obviously reacted with about $1.0\times10^{-7}$ moles of $\ce{[OH^-]}$. But this amount is so small compared to the 1.0 moles of $\ce{[H^+]}$ that it is insignificant. 


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