# Confusion on ion product of water

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.

• The acid provides additional $\ce{H+}$. Hydroxyls present in water react with this to form back more water. – Buck Thorn Nov 4 '19 at 22:08
• Two words: 'Detailed balance' would be a good place to start... – Jon Custer Nov 4 '19 at 23:26
• @BuckThorn Yes that's very agreeable, but how does the concentration of hydroxyl ions increases when we add base to water? – Knight Nov 5 '19 at 7:35
• 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. – Knight Nov 5 '19 at 16:21

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 $$K_w$$.

• 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. – Knight Nov 5 '19 at 16:04
• 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. – Ezze Nov 5 '19 at 16:10
• 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 $pH + pOH = 14$. – Ezze Nov 5 '19 at 16:12
• 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 . – Knight Nov 5 '19 at 16:19
• 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 $\ce{NH3}$. – Ezze Nov 5 '19 at 16:23

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 $$K_\mathrm{w}$$

$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$

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

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 $$[\ce{H+}][\ce{OH-}] = 10^{-14}\,\mathrm{M}$$.

• 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? – Knight Nov 4 '19 at 16:34
• No ! The dissociation of H2O is hindered by both the presence of an acid and of a base. – Maurice Nov 4 '19 at 17:58