The auto-ionization of H2O equilibrium

[Roq]

I'm trying to conceptually understand a question that I was wondering about while reading about Acid and Base Equilibria.

Since a strong acid such as HCl completely reacts to give Cl- and the hydronium ion, and Cl- is then inert, then the only equilibrium going on is Kw. Kw = KbKa = [OH-][H+] = 1x10^-14...

I always thought that when OH- or H+ were produced, the Kw would balance itself like other reactions, and in this case it would mean the reaction going to the reactant direction.

Since the equilibrium would be restored by reacting [OH-] and [H+], they would both decrease on a one to one basis, and it could not just decrease the [OH-] without in turn decreasing [H+] at the same time.

That seems logical to me, but given the way these equations are calculated (at least in my class), like in the case of HCl, you would have an equal [HCl] and [H+] concentration, and there is no subtraction to account for the equilibrium of Kw. While this may not matter if you are dealing with 1 M of HCl, if your concentration is very small, like around 1x10^-7, then a large percentage of it would be subtracted as Kw is restored.

Perhaps it is irrelevant because there is little use for acids and bases at such tiny concentrations, but is this correct?

 

[chemisttree]

If the concentration of HCl were 1X10^-7, the expression for Kw becomes:

Kw = [OH-][H+(from HCl) + H+(from autoionization)]

Let x = [H+(from autoionization)] We see that [OH-] must also be equal to x from the equation

HOH <---> H+(from autoionization) + OH-(from autoionization)

Substituting x into the Kw expression gives us:

Kw = x([H+](from HCl) + x)
Kw = x^2 + [H+](from HCl)x
0 = X^2 + [H+](from HCl)x - Kw

Solving this quadratic equation gives us x = 6.1803X10^-8. This is the new concentration for [OH-]. The total concentration for H+ is 1.62 X 10^-7. The pH is 6.79... slightly on the acid side as you would expect.

Kw does balance itself as in other reactions. If [H+] were 1M (pH=0), the concentration of [OH-] would become 1X10^-14. A pretty small number.

 

[Borek]

First of all - it is not true that strong acids dissociate completely. Their strength can be described by dissociation constant as well, just its value is huge (like 10^7 for HCl).

As for pH of the strong acid - look at these two pages:

general approach to pH calculation

calculation of pH of strong acid

 

[Roq]

Thank you, guys.

 

[GCT]

I'm trying to conceptually understand a question that I was wondering about while reading about Acid and Base Equilibria.

Since a strong acid such as HCl completely reacts to give Cl- and the hydronium ion, and Cl- is then inert, then the only equilibrium going on is Kw. Kw = KbKa = [OH-][H+] = 1x10^-14...

I always thought that when OH- or H+ were produced, the Kw would balance itself like other reactions, and in this case it would mean the reaction going to the reactant direction.

Since the equilibrium would be restored by reacting [OH-] and [H+], they would both decrease on a one to one basis, and it could not just decrease the [OH-] without in turn decreasing [H+] at the same time.

That seems logical to me, but given the way these equations are calculated (at least in my class), like in the case of HCl, you would have an equal [HCl] and [H+] concentration, and there is no subtraction to account for the equilibrium of Kw. While this may not matter if you are dealing with 1 M of HCl, if your concentration is very small, like around 1x10^-7, then a large percentage of it would be subtracted as Kw is restored.

Perhaps it is irrelevant because there is little use for acids and bases at such tiny concentrations, but is this correct?

Yes, in some cases, the auto-ionization of water is taken into account in finding the pH value, in the case of HCl, the pH is going to be accounted for by the extra "H+" that has been contributed by the HCl. With very small concentrations of HCl, Kw needs to be employed to obtain a significantly more accurate pH value.