The ion product of water is defined as $$ K_w = [\ce{H+}][\ce{OH-}] $$ and in pure water $ [\ce{H+}] = [\ce{OH-}]= 1.0 \times 10^{-7}$ and $ K_w = 1.0 \times 10^{-14}$. It is said that if we dissolve some acidic substance in water then $[\ce{H+}]$ will increase and $[\ce{OH-}]$ will decrease and the value of $K_w$ remains constant. I understand that $[\ce{H+}]$ increases because acids donate $\ce{H+}$ (according to Brønsted definition of acids) but how $[\ce{OH-}]$ decreases. I mean a water molecule i.e. $\ce{H2O}$ will certainly break into $\ce{H+}$ and $\ce{OH-}$ and to this $\ce{H+}$ our acid has added it's part and hence it's concentration has increased but why the value of $[\ce{OH-}]$ gets lowered from $1.0 \times 10^{-14}$ , if water molecule were to ionize then whenever $\ce{H+}$ gets formed simultaneously we would get $\ce{OH-}$.

I can cite a common example problem

The concentration of $\ce{OH-}$ ions in a certain household ammonia cleansing solution is $0.0025$. Calculate the concentration of $\ce{H+}$ ions.

We can solve this problem by using the equation $$ K_w = 1.0 \times 10^{-14}$$ $$ [\ce{H+}] [\ce{OH-}] = 1.0 \times 10^{-14}$$ and if put the value of $[\ce{OH-}]$ in the above equation and solve $[\ce{H+}]$, then we would get $ 4.0 \times 10^{-12}$. Here we observe that the value of $[\ce{H+}]$ has decreased from the it's original value in pure water, well this is understandable because ammonia being a base would consume $\ce{H+}$ but how the value of $\ce{OH-}$ has increased.

I want to know that how the chemical reaction can cause the increase or decrease of $\ce{OH-}$ when each molecule of $\ce{H2O}$ always going to yield one $\ce{H+}$ and one $\ce{OH-}$ always.

Thank you. Any help will be much appreciated.

  • $\begingroup$ The acid provides additional $\ce{H+}$. Hydroxyls present in water react with this to form back more water. $\endgroup$
    – Buck Thorn
    Nov 4, 2019 at 22:08
  • $\begingroup$ Two words: 'Detailed balance' would be a good place to start... $\endgroup$
    – Jon Custer
    Nov 4, 2019 at 23:26
  • $\begingroup$ @BuckThorn Yes that's very agreeable, but how does the concentration of hydroxyl ions increases when we add base to water? $\endgroup$ Nov 5, 2019 at 7:35
  • 1
    $\begingroup$ I just want to ask cordially that how my question is a homework problem, I mean it may be a trivial question for well educated and researchers over here but I can't see how it is homework problem. $\endgroup$ Nov 5, 2019 at 16:21

3 Answers 3


It is not a chemical problem, but a trivial mathematical problem.

If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.

The key part is to understand that chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like

$$\ce{H2O <=> H+ + OH-}$$

where the rate of ion creation equals the rate of ion recombination.

If there is an excess or deficit of either of $\ce{H+}$ or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.

The ammonia reacts :

$$\ce{NH3 + H+ <=>> NH4+}$$

That creates deficit of $\ce{H+}$. As consequence, dissociation

$$\ce{H2O -> H+ + OH-}$$

Is faster then recombination

$$\ce{ H+ + OH- -> H2O}$$

An alternative reaction mechanism is ammonia reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.

The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water $K_\mathrm{w}$


$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$


Yes. If some acid is added to pure water, $[\ce{H+}]$ increases but $[\ce{OH-}]$ decreases. It means that:

  1. in pure water, enough $\ce{H2O}$ will break into $\ce{H+}$ and $\ce{OH-}$.
  2. in the presence of an acid, a smaller amount of water will break into $\ce{H+}$ and $\ce{OH-}$. The presence of an acid prevents $\ce{H2O}$ from being dissociated.

For example, if you add $10^{-7}\mathrm{mol}$ of acid in 1 liter pure water, the concentration $[\ce{H+}]$ is the sum of the added $\ce{H+}$ ($10^{-7}\,\mathrm{M}$) and of the smaller amount of water being broken. The concentration of $\ce{OH-}$ is decreased in the same way. Calculation shows that only $0.618\,\mathrm{mol}$ water are broken into $\ce{H+}$ and $\ce{OH-}$, if $10^{-7}\,\mathrm{mol}$ acid is added.

With this value, the final concentrations are: $[\ce{H+}] = 1.618 10^{-7}\,\mathrm{M}$, and $[\ce{OH-}] = 0.618 10^{-7}\,\mathrm{M}$. You may check that the product $[\ce{H+}][\ce{OH-}] = 10^{-14}\,\mathrm{M}$.

  • $\begingroup$ Okay. I have understood that the dissociation of $H_2O$ gets hindered due to the presence of an acid but speeds up in presence of a base, am I correct? $\endgroup$ Nov 4, 2019 at 16:34
  • $\begingroup$ No ! The dissociation of H2O is hindered by both the presence of an acid and of a base. $\endgroup$
    – Maurice
    Nov 4, 2019 at 17:58

Another way to look at the problem:

If we add $\ce{H+}$ to the water through adding acid, then some of the $\ce{H+}$ would just remain in the solution as is, and some of them would react with $\ce{OH-}$ in the reaction $\ce{H+ + OH- -> H_2O}$. How much of the added $\ce{H+}$ reacts? To calculate this, you need to know that $\ce{[H+][OH-] = K_w}$ remains constant. So while $\ce{H+}$ is increased, $\ce{OH-}$ is decreased through $\ce{H+ + OH- -> H_2O}$.

If we were to add $\ce{OH-}$ to the water through bases, the same thing would happen: a portion would react with $\ce{H+}$ and a portion would remain in the solution, and it will happen in such a way that is dictated by the constant value of $K_w$.

  • $\begingroup$ Thank you, your answer is very clear and acceptable. I just want to correct you (or you can correct me if I'm wrong) that base doesn't add $OH^-$ if we use the definition of BrØnsted base, it just consumes $H^+$. So, I think when we add a base to water $H^+$ gets consumed and hence equilibrium shifts to the right side and consequently we get more $OH^-$ . But I expect a more clear explanation from you, just the same way you have explained the acid problem. $\endgroup$ Nov 5, 2019 at 16:04
  • $\begingroup$ You can look at the problem from any side you want. In my comment I argued that adding let's say $\ce{NaOH}$ to water increases the $\ce{OH-}$ concentration, since it dissociates into $\ce{Na+}$ and $\ce{OH-}$ . On the other hand this means that $\ce{H+}$ is consumed - exactly because of the constant value of $K_w$. Bronsted's theory always looks at the concentration of $\ce{H+}$ (hence the definition of $pH$), but it is equally reasonable to think about $\ce{OH-}$. It is the two sides of the same coin, and the connection between the two is exactly the $K_w$ equation in your question. $\endgroup$
    – user23638
    Nov 5, 2019 at 16:10
  • $\begingroup$ By the way when I used to teach chemistry calculations I liked the idea of introducing the quantity $pOH$, which is an analogue of $pH$, defined by $pOH = - lg[\ce{OH-}]$. Here the $K_w$ value means that $pH + pOH = 14$. $\endgroup$
    – user23638
    Nov 5, 2019 at 16:12
  • $\begingroup$ Is it true that all acids will always gonna release $\ce{H+}$ and all bases always release $\ce{OH-}$. Because the example problem which I have cited, $\ce{NH3}$ can't release $\ce{OH-}$ on its own, it need water for that. Please explain . $\endgroup$ Nov 5, 2019 at 16:19
  • $\begingroup$ Release is not true. What is true is that acids increase the $\ce{H+}$ concentration and bases decrease the $\ce{H+}$ concentration (or, equally speaking, increase the $\ce{OH-}$ concentration). In the case of $\ce{NH_3}$, as seen from Poutnik's answer, we have $\ce{NH3 + H+ -> NH_4+}$. This reaction decreases the $\ce{H+}$ concentration, but since $[\ce{H+}][\ce{OH-]}$ is constant, tthis means that $\ce{OH-}$ is increased. But not from releasing from $\ce{NH3}$. $\endgroup$
    – user23638
    Nov 5, 2019 at 16:23

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