0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The acid provides additional $\ce{H+}$. Hydroxyls present in water react with this to form back more water. $\endgroup$ – Buck Thorn Nov 4 '19 at 22:08
  • $\begingroup$ Two words: 'Detailed balance' would be a good place to start... $\endgroup$ – Jon Custer Nov 4 '19 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? $\endgroup$ – Knight Nov 5 '19 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. $\endgroup$ – Knight Nov 5 '19 at 16:21
1
$\begingroup$

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 $K_w$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Knight Nov 5 '19 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. $\endgroup$ – Ezze Nov 5 '19 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 $pH + pOH = 14$. $\endgroup$ – Ezze Nov 5 '19 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 . $\endgroup$ – Knight Nov 5 '19 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 $\ce{NH3}$. $\endgroup$ – Ezze Nov 5 '19 at 16:23
3
$\begingroup$

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 $K_\mathrm{w}$

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

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

| improve this answer | |
$\endgroup$
2
$\begingroup$

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Knight Nov 4 '19 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 '19 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.