Acid-base equilibrium
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For many years chemists have classified substances into acids or bases. Lavoisier thought that acidic substances owed their behavior to the presence of oxygen. But it was Humphry Davy in 1810 who showed that the element that all acids have in common is hydrogen.
The first acid-base theory was developed by Svante Arrhenius in 1884, explaining acidity as the ability of a substance to dissociate by giving up protons, and basicity as the ability to give up hydroxide ions.
In 1923, Bronsted and Lowry proposed a theory that the acid was a proton donor and the base was an acceptor, so that acid-base reactions occurred via conjugate acid-base pairs.
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Arrehius's acid-base theory stems from his studies on electrolytes. Arrhenius proposed that an electrolytic substance dissociates in water forming ions, furthermore he classified 
strong and weak electrolytes depending on whether their dissociation is complete or incomplete.
Based on these ideas, he explained that an acid, such as HCl, completely dissociates in water (strong acid) generating protons, responsible for its acidity.
$HCl + H_2O \rightarrow Cl^- + H_3O^+$
Bases, for their part, when dissolved in water, produce hydroxide ions (OH - ), responsible for their basicity. So sodium hydroxide (NaOH) completely dissociates in aqueous solution, forming sodium cations and hydroxide ions that make it a strong base.
$NaOH \rightarrow Na^+ + OH^-$
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In 1923, Bronsted and Lowry independently proposed a theory to explain the acid-base behavior of substances, in which acids are proton donors while bases are proton acceptors.
This theory naturally explains, unlike Arrhenius, the basic behavior of ammonia.
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$
Water acts as an acid by donating a proton to ammonia, which in turn acts as a base. Water becomes its conjugate base; the hydroxide ion. While ammonia is converted to its conjugate acid; the ammonium ion.
This proton transfer between water and ammonia releases hydroxide ions into the medium, responsible for the basicity of the solution.
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Conductivity measurements indicate that pure water is partially ionized. Ions are generated when some water molecules act as acids by donating protons to other water molecules that act as bases, according to the following equilibrium.
$H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$
This is a reversible reaction highly shifted to the left. The amounts of oxonium and hydroxide ions present in the medium are very small.
The equilibrium constant for this reaction is called the ionic product of water, $K_w$
$K_w=[H_3O^+][OH^-]=10^{-14}$
In pure water the concentrations of both ions are the same: $[H_3O^+]=[OH^-]=10^{-7}$
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A strong acid is one that completely dissociates, although we must bear in mind that this statement would not be true in highly concentrated solutions. An example of a strong acid is hydrochloric acid.
$HCl +H_2O\rightarrow Cl^{-}+H_3O^+$
Note that I have used an arrow to indicate the full displacement of the equilibrium to the right.
With this type of acid, the autoionization of water is negligible and the only source of protons is the acid (valid for solutions that are not extremely dilute).
From the above it follows that: $[HCl]=[H_3O^+]$, where $pH=-log[H_3O^+]$
Let's see an example: Calculate the pH of a 0.1 M HCl solution.
Being a strong acid $[HCl]=[H_3O^+]=0.1\;M$
pH=-log(0.1)=1
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Alkali and alkaline earth metal hydroxides are the most common strong bases. These bases release hydroxide groups into the aqueous medium, responsible for their basicity. Since the hydroxide ions generated by the autoionization of water are negligible compared to those generated by the base (except when the base is very dilute), we can consider that the base is the only source of hydroxide ions in the medium, and therefore both $[OH^-]=[base]$, where $pOH=-log[OH^-]$ and $pH=14-pOH$
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In the previous sections we calculated the pH in solutions of strong acids and bases. In these cases, the concentration of protons or hydroxide ions coincides with the concentration of the acid or the base, and we only have to take the logarithm of said concentration to obtain the pH or pOH. But we are surprised that a $10^{-8}M$ HCl solution has a basic pH of 8. Logic makes us think that said solution should have a pH slightly below 7, although very close to this value .
The explanation for this "surprise" lies in neglecting the autoinoculation of water, since it contributes more protons to the medium than the acid itself. Calculation of pH in this situation should be done as follows.
$HCl+H_2O\rightarrow \underbrace{H_3O^+}_{10^{-8}}+ \underbrace{Cl^-}_{10^{-8}}$
$H_2O+H_2O\rightleftharpoons \underbrace{H_3O^+}_{x+10^{-8}} +\underbrace{OH^-}_{x}$
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A weak acid is one that partially ionizes in water. A typical example is acetic acid.
$CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+$
The ionization reaction does not shift completely to the right as in the case of strong acids, establishing an equilibrium between the dissociated species and the undissociated species that is given by the acidity constant.
\begin{equation}K_a=\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}\end{equation}
From an initial concentration of acid, and using the acidity constant, we will be able to determine the concentration of protons in the medium and consequently the pH of the solution.
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This is a situation analogous to the calculation of pH with weak acids that we discussed in the previous point. Weak bases in solution do not completely dissociate as strong bases do, so an equilibrium is established between the base and its dissociated species. This equilibrium is evaluated through the basicity constant, which will allow us to calculate the concentration of free hydroxide ions in the medium. Once this concentration is found, we proceed to calculate the pOH, and with its relationship with the pH we obtain the latter.
Let's see an example: Calculate the pH of a 0.0025 M solution in methylamine. ($K_b=4.2x10^{-4}$).
We begin by writing the dissociation equilibrium of methylamine:
$CH_3NH_2 + H_2O \rightarrow CH_3NH_3^+ + OH^-$
Initial 0.0025 -------- ---------
Equilibrium 0.0025-x x x
Taking the equilibrium composition to the constant:
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When a weak acid is very dilute, it is necessary to take into account the autoionization of water. We are going to write the two equilibria that take place in solution:
$H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$
x x
$HA +H_2O\rightleftharpoons H_3O^+ + A^-$
M-y y y
Both M, x and y are expressed in molarity. It must be taken into account that the concentration of protons in the medium is given by x+y.
Read more: Calculation of the pH of a very dilute solution of a weak acid
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So far we have studied acids that had only one ionizable hydrogen (monoprotic acids). However, there are acids that have several ionizable hydrogens, called polyprotic acids. Some examples are acids: $H_3PO_4 \;H_2SO_4$
The $H_3PO_4$ molecule is a triprotic acid, with three ionizable hydrogens. The acidity of the hydrogens decreases rapidly, the first being much more acid than the second, and the latter than the third. The reason is that as we remove hydrogens, the molecule acquires a negative charge and the conjugate base is increasingly unstable (strong).
Let's see the three equilibrium in the ionization of phosphoric acid:
$H_3PO_4 + H_2O \rightleftharpoons H_2PO_4^- +H_3O^+$ $K_{a1}=7.1x10^{-3}$
$H_2PO_4^- +H_2O\rightleftharpoons HPO_4^{2-} +H_3O^+$ $K_{a2}=6.3x10^{-8}$
$HPO_4^{2-}+H_2O\rightleftharpoons PO_4^{3-} +H_3O^+$ $K_{a3}=4.2x10^{-13}$
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Sulfuric acid is a diprotic acid, with the special characteristic of presenting a first complete dissociation. The pH of a sulfuric acid solution is given, approximately, by the protons generated in the first dissociation since the second presents little extension.
Let's see the ionization reactions of sulfuric acid:
$H_2SO_4 +H_2O\rightarrow HSO_4^- +H_3O^+$
$HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+$ $k_a=1.1x10^{-2}$
As can be seen, the first reaction is completely shifted to the right and its equilibrium constant is not shown since it is very large. On the contrary, the second reaction does not totally displace and we must evaluate its equilibrium as a function of the constant, as well as the protons generated in the first reaction.
Let us now calculate the pH of a 0.5 M solution of sulfuric acid.
$H_2SO_4 +H_2O\rightarrow HSO_4^- +H_3O^+$
--- --- 0.5 0.5
$HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+$
0.5 - x x 0.5 + x
We evaluate the constant of this second equilibrium:
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So far we have seen how the addition of acids or bases to water modifies its pH. However, salts can also change the pH of water by hydrolyzing. Hydrolysis is a process by which the cation or anion of a salt, provided it comes from a weak base or weak acid, reacts with water to generate its conjugate base or acid.
Let's see some examples:
- Sodium acetate dissociates in water, generating acetate anions and sodium cations. Sodium cations coming from a strong base (NaOH) do not hydrolyze in water, but acetate anions coming from a weak acid ($CH_3COOH$) undergo the following transformation:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
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In the previous section I showed that salts hydrolyze in the aqueous medium, generating acidic or basic media. Now we will calculate the pH of a 0.5 M aqueous solution of sodium cyanide.
The cyanide anion comes from a weak acid (HCN) and undergoes the following hydrolysis reaction:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
As can be seen, the hydrolysis reaction generates hydroxide ions and will produce a basic pH.