# When we dilute a weak acid, pH increases but dissociation increases

When water is added to a weak acid like ethanoic aicd, number of ethanoic acid molecules that dissociate increases however, pH increases (less acidic)... but why?

It's simply due to dilution.

If it's true that adding water increases dissociation (@Gerard has done a nice effort here), concentration of $\ce{H3O+}$ inevitably decreases. As a consequence, $pH$, defined as $-log[\ce{H3O+}]$, increases.

Start with $1$ liter acetic acid concentration $1M$

$\ce{[H3O+]}=\sqrt{K_c}=4.22 \times 10^{-3}\,M$

that corresponds to $4.22 \times 10^{-3}\,mol$ of $\ce{H3O+}$

If you double the volume ($2$ liters):

$\ce{[H3O+]}=\sqrt{K_c/2}=2.98 \times 10^{-3}\,M$

that corresponds to $5.96 \times 10^{-3}\,mol$ of $\ce{H3O+}$.

The number of moles is higher, but concentration is less: dilution does play a role here.

• You should use $mol\cdot L^{-1}$ instead of $M$ for concentrations. This might not effect a compound with 1:1 stoichiometry but all others. It is also lab shorthand. – Martin - マーチン Mar 31 '14 at 8:16
• @Martin are you talking of normality ($N$) ? In that case stoichiometry plays a role. On the other hand, $M$ and $mol\,L^{-1}$ can be alternatively used: CRC Handbook Chemistry and Physics 85th 2-51 or en.wikipedia.org/wiki/Molar_concentration#Units – mannaia Mar 31 '14 at 11:06
• I understand, and you are right in many cases. (My previous comment might seem a little harsh, I appologize, It was only meant as a sugesstion.) For example, if you have a $\ce{CaCl2}$ solution with $1M$, the ion concentrations ($\ce{Ca^{2+},Cl-}$) are different ($1mol\cdot L^{-1}$ and $2mol\cdot L^{-1}$) and to avoid any confusion I would suggest sticking to SI units. It also makes calculations a little bit more obvious. Normality on the other hand is only confusing - I personally do not know how to correctly use that concept. But that is a different question :D – Martin - マーチン Apr 1 '14 at 1:43

Let's first examine the chemical reaction between our acids and bases. Ethanoic acid, commonly referred to as acetic acid reacts with our base, water. Water is amphiprotic, in that it can act as both a base and acid.

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

However, the acetate ion undergoes hydrolysis with water which now instead acts as an acid, reforming acetic acid as well has hydroxide ions:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

Acetic acid is weak acid, however, the conjugate base, acetate, is a strong base/salt. Water acts as a strong base, however the hydronium ion is a weak acid, so we can except the pH to increase.

What many people often forget to realize is that once an acid/base reaction occurs, a salt is created along with water. However, a hydrolysis reaction between the salt and water can occur.

Taken from Chemical Principles: The Quest for Insight by Atkins et. al on page 494 in the tenth edition:

Calcium acetate, $\ce{Ca(CH3CO2)2(aq)}$ is ued in medicine to treat patients with a kidney disease that results in high levels of phosphate ions in the blood. The calcium binds with the phosphates so that they can be excreted. If you are using calcium acetate for this purpose, it is important to know the pH of the solution to avoid complications in treatment.

Say we have a 0.15M solution of $\ce{Ca(CH3CO2)2(aq)}$.

Since the conjugate base of acetic acid is strong, we can expect the pH to increase.

Our expression can be written as this:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

Calcium is a spectator ion and thus can be ignored.

$$K_b = \ce{\frac{[CH3COOH][OH-]}{[CH3COO-]}}$$

Constructing our RICE table:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

\begin{array} {|c|c|c|c|c|} \hline \text{Initial conc.} & 0.30 & - & 0 & 0 \\ \hline \text{Change conc.} & -x & - & +x & +x\\ \hline \text{End conc.} & 0.30 - x & - & x & x \\ \hline \end{array}

$$K_b = 5.6*10^{-10} = \frac{[x][x]}{[0.30-x]}$$

$$x = 1.3 * 10^{-5}$$

$$\text{pOH} = -log([\ce{OH-}])$$

$$\text{pH} = 14 - \text{pOH} = 14 - 4.89 = 9.11$$

• "What many people often forget to realize is that once an acid/base reaction occurs, a salt is created along with water. However, a hydrolysis reaction between the salt and water can occur." That is not true. – Martin - マーチン Apr 1 '14 at 6:48
• @Martin So the acetate ion does not undergo hydrolysis with water? – Jun-Goo Kwak Apr 1 '14 at 12:20
• Of course it does, but acid base reactions do not always form salts - that statement is much too general. The neutralisation of Ethanoic acid with Sodium hydroxide yields solvatised ions, which may be expressed as a salt due to incomplete dissociation (it is still a solution): $$\ce{H3C\bond{-}COOH + NaOH_{(aq)} <=> H3C\bond{-}COO- + H2O + Na+_{(aq)}}$$ Hydrochloric acid and Sodium hydroxide just make a sodium chloride solution (no salt): $$\ce{HCl_{(aq)} + NaOH_{(aq)} -> H2O + Na+_{(aq)} + CL-_{(aq)}}$$ If you get rid of the water it will form the crystal. – Martin - マーチン Apr 2 '14 at 2:08
• @Martin Yes, the salt can dissociate into solvatised ions. That is understood. The net ionic is just a proton donated to a hydroxide to form water. I didn't know that that knowledge is too "general" and as such, will make my answers more specific. – Jun-Goo Kwak Apr 2 '14 at 2:16