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I know in neutral solution, $\ce{[H+]}= \ce{[OH-]}= \pu{1.0 x 10^-7}$.

However, let’s say I add an acid like HCl. This will increase $\ce{[H+]}$, therefore it’s no longer $\pu{1.0 x 10^-7}$, but is greater.

So, $\ce{[OH-]}$ would need to decrease to keep $K_w$ at a constant $10^{-14}$, reestablishing equilibrium.

However, how does this $\ce{[OH-]}$ decrease? What does it react with to decrease its concentration? At first I thought it could react with $\ce{[H+]}$ to make water, but that’d be decreasing $\ce{[H+]}$, which doesn’t make any sense since the solution should be acidic if HCl was added.

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    $\begingroup$ Note you've forgotten to include the molar units for the concentrations. $\endgroup$
    – Buck Thorn
    Oct 25, 2023 at 8:54

4 Answers 4

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Dissociation equilibrium in water is attained when the rates of the following forward and reverse reactions are equal (all species aqueous of course):

$$\ce{H2O ->[k_f] H+ + OH-} $$ $$\ce{H+ + OH- ->[k_r] H2O}$$

The water molecules dissociate with a regular frequency (you can think of each water molecule having a lifetime before it falls apart), and the collision rate between oppositely charged ions depends to a first degree on their concentrations. If you have more $\ce{H+}$ then you need less $\ce{OH-}$ for a "productive" collision resulting in formation of a water molecule, and vice-versa.

The forward an reverse rates become equal when

$$k_f [\ce{H2O}] = k_r[\ce{H+}][\ce{OH-}] \tag{1}$$

In practice activities are used instead of concentrations but for a solution dilute in the ions using concentrations expressions for the ions is ok, as the encounter rate between ions can be assumed to vary linearly with concentration for dilute solutions.

This leads to an expression for the dissociation constant of water:

$$K = \frac{k_f}{k_r} = \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]}$$

Strictly speaking activities are used and $K_w$ is defined as:

$$K_w = \frac{a_\ce{H+} a_\ce{OH-}}{a_\ce{H2O}} $$

At low concentrations (approx neat water) $a_\ce{H2O}=1$ and the activities can be replaced by concentrations such that

$$K_w \approx [\ce{H+}][\ce{OH-}] \tag{2}$$


I browsed Ref. 1 to double-check my answer. I realized my exclusion of hydronium was asking for trouble. Strict definitions state the reaction as $\ce{2H2O <=>[K_w] H3O+ + OH-}$. But I prefer to leave the answer as is, as I did write aqueous, and because the definition of pH is closely related to this concept. But the definition of $K_w$ as constant is still imperfect. It is not truly a constant when you start adding other stuff to water. $K_w$ defines the autoionization constant of pure water for a given p and T. But if dilute in other added ions, then the prediction of simple kinetic theory, Eq. 1, leads to the conclusion that Eq. 2 with $K_w$ assumed constant can be used.


Note also: In practice when you measure pH you are strictly-speaking not measuring the concentration of $\ce{H+}$ but rather its activity.


References

Andrei V. Bandura; Serguei N. Lvov. The Ionization Constant of Water over Wide Ranges of Temperature and Density In Special Collection: International Water Property Standards. J. Phys. Chem. Ref. Data 35, 15–30 (2006) https://doi.org/10.1063/1.1928231

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    $\begingroup$ This explanation is absolutely correct. But it does not give an answer to the question posed by Maria. $\endgroup$
    – Maurice
    Oct 25, 2023 at 9:32
  • $\begingroup$ @Maurice Well, if you read to the end: "When using activities K becomes a concentration independent constant (concentration dependence of the rate constants being incorporated into the activities)". In addition the beginning of the answer attempts to explain how the concentration of different species is affected by collisions and water dissociation, which is very pertinent to clarifying the problem. $\endgroup$
    – Buck Thorn
    Oct 25, 2023 at 9:47
  • $\begingroup$ @Maurice BTW your answer addresses how one computes the pH (or how pH depends on concentration of added acid) but not why Kw is constant. $\endgroup$
    – Buck Thorn
    Oct 25, 2023 at 9:52
  • $\begingroup$ You are right. But there is a strange discrepancy between the title (Is Kw constant ?) and the text itself (How will [$\ce{OH-}$] decrease ?). I have only answered the question in the text. $\endgroup$
    – Maurice
    Oct 25, 2023 at 14:00
  • $\begingroup$ @Maurice Granted. I guess the answers are complementary in a way. $\endgroup$
    – Buck Thorn
    Oct 25, 2023 at 15:22
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If you add $\ce{HCl}$ in water, the amount of $\ce{H+}$ increases of course. And part of this supplementary $\ce{H+}$ ions are used to destroy $\ce{OH-}$ ions from pure water. This amount of $\ce{H+}$ required to do this job must be smaller than or equal to $10^{-7}$ M. It is rather small. It is so small that it can be neglected with respect to the amount brought by the supplementary $\ce{HCl}$, which can give a concentration having any numerical value bigger than (let's say)... $10^{-3}$ M or $10^{-2}$ M.

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The hydroxide ions will react back indeed, albeit with hydronium ions (not $\ce{H^+}$), but because there are only $10^{-7}$ mol/l hydroxide ions in solution, it can also react with a maximum amount of $10^{-7}$ hydronium ions $\ce{H_3O^+}$, therefore basically not changing the HCl‘s much,much higher own concentration.

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Kw is not constant. Kw = Keq [H2O]. As the activity of water changes Kw changes. The activity of water is reasonably constant in dilute aqueous solutions. Elsewhere things get complicated, changes in water activity have effects. This is noticeable in cooking and the behavior of alcohol-water mixes and most likely in cellular chemistry.

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  • $\begingroup$ This answer is true,but addresses the effects few levels below the trivial basics the OP asks about. $\endgroup$
    – Poutnik
    Oct 26, 2023 at 20:56

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