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In my textbook, a footnote says:

In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases

This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:

$$\ce{HA (aq) + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$

Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me.

Delving into the math a bit, we can see that

$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$

Since the concentrations of the two products are equal,

$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$

Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions.

Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?

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    $\begingroup$ Haven't you pretty much answered your own question? This looks like a great answer to me. The intuition is Le Chatelier, and you proved the intuition with math. Win! $\endgroup$ – user467 Jan 4 '14 at 18:45
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Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:

$\hskip2in$ dependence of the pH of an acid solution with pK_a 5

Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$

Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)

$\hskip2in$ composition of an acid base system depending on the given pH

Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$

One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.

So for me this is no big intuition thing but rather a small sequence of actions.


Lets say it in math:

The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$


One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.

$\hskip2in$ enter image description here

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    $\begingroup$ Would you please elaborate how you got that equation for disassociation constant? $\endgroup$ – Mockingbird May 22 '17 at 2:19
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    $\begingroup$ The first equation for $\alpha $? $\endgroup$ – pH13 - Yet another Philipp May 22 '17 at 6:06
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    $\begingroup$ Yeas, I meant that. $\endgroup$ – Mockingbird May 22 '17 at 7:27
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    $\begingroup$ I’m not sure if such a question exists on chem.se. But if not, it would be nice for others, if you’d ask it. $\endgroup$ – pH13 - Yet another Philipp May 22 '17 at 8:27
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    $\begingroup$ Your question didn’t sound like you know the solution but that you want to get an answer. But as you say, that it does not look like this, you seem to know the answer. So, please, how is it? And btw, that "I am telling you to ..."-formulation is quite rude, as you can ask but not tell me to do sth. $\endgroup$ – pH13 - Yet another Philipp May 22 '17 at 12:44
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Intuition behind this concept is thermodynamically based.

If you think of $K_a$ as a thermodynamically favorable constant that describes the analyte concentrations associated with acid dissociation then it is not too difficult to reconcile the increase in $\ce{H+}$ ions in solution as a necessary means to maintain a thermodynamically favorable state when water is added to solution.

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This is one of the common problems of ionic equilibrium. Generally questions are asked to calculate the $\mathrm{pH}$ of a diluted weak acidic solution and if the $\mathrm{pH}$ calculation turns out to be $7$ or more than that, we have to add the contribution of water also. I am giving you an example that will resolve your matter completely:

Calculate the $\mathrm{pH}$ of $\pu{1.0E-8 M}$ solution of $\ce{HCl}$.

For the reaction $$\ce{HCl + H2O <=> H3O+ + Cl-}\tag1\label{hcl-dissociation},$$

if you calculate the pH as $$\text{pH} = -\log(10^{-8}),\tag2$$ then $\text{pH}=8$. Now you can see that the $\mathrm{pH}$ of $\ce{HCl}$ can not be $8$, it has to be less than $7$, therefore you have to include the contribution of water.

Consider \begin{align} \ce{2H2O &<=> H3O+ + OH-}\tag3\label{selfionisation}\\ K_\mathrm{w} &=\ce{[OH-][H3O+]}\\ K_\mathrm{w} &=10^{-14} \end{align}

Therefore the $\ce{H3O+}$ or $\ce{H+}$ concentration increases from:

  1. Ionization of $\ce{HCl}$ dissolved \eqref{hcl-dissociation},
  2. Self ionization of water \eqref{selfionisation}.

So in very dilute solutions like this solution we have to consider the contribution of both sources of $\ce{H3O+}$; this also explains why the concentration of hydronium ions increases upon dilution of a weak acidic solution.

now going back to problem

\begin{align} \text{let } x = [\ce{OH-}] &= [\ce{H3O+}]\\ [\ce{H3O+}] &= 10^{-8} + x\\ K_\mathrm{w} &= (10^{-8} + x)(x) = 10^{-14}\\ 0 &= x^2 +10^{-8}x -10^{-14}\\ [\ce{OH-}] &= x = \pu{9.5E-8}\\ \mathrm{pOH}&=7.02\\ \mathrm{pH}&=6.98 \end{align}

Now you can see that we get $\mathrm{pH} = 6.98$ after adding the contribution of water, otherwise we would have gotten the absurd answer of $\mathrm{pH}=8$, which can not be true.

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