Acids and bases can be classified as either monoprotic or polyprotic. Monoprotic acids and bases only release/accept one \(H^{+}\) ion, while polyprotic acids and bases release/accept more than one. This article will focus on polyprotic acids and bases, including reactions between polyprotic acids and bases, and how properties of polyprotic acids and bases affect pH.
Polyprotic acids will release multiple hydrogen ions. As a result, every molecule of all polyprotic acids must have at least two hydrogen ions. The stability of the structure of the molecule sometimes plays a role in how many hydrogen ions a polyprotic acid molecule will release. Many polyprotic acids will turn into common polyatomic anions after releasing between two and four hydrogen ions. The following is a list of pairs of these acids and the anions they form:
$$H_2CO_3 \Rightarrow CO_3^{2-}$$ $$H_2SO_4 \Rightarrow SO_4^{2-}$$ $$H_2SO_3 \Rightarrow SO_3^{2-}$$ $$H_3PO_4 \Rightarrow PO_4^{3-}$$ $$H_3PO_3 \Rightarrow PO_3^{3-}$$ $$H_4SiO_4 \Rightarrow SiO_4^{4-}$$ $$H_4SiO_3 \Rightarrow SiO_3^{4-}$$
As a side note, the first polyprotic acid listed, carbonic acid, will typically decompose into water and carbon dioxide instead of two hydrogen ions and a carbonate ion, but the shown transformation is still theoretically possible.
Example 1: Show the steps that a molecule of sulfuric acid will undergo to become a sulfate ion by means of chemical equations.
Solution: Sulfuric acid is \(H_2SO_4\). The table above suggests that
$$H_2SO_4 \rightarrow 2H^{+} + SO_4^{2-}$$
As a result, sulfuric acid is a diprotic acid, and there will be two hydrogen ion releases, each one constituting a step in the overall reaction. Thus we get these chemical equations:
$$H_2SO_4 \rightarrow H^{+} + HSO_4^{-}$$ $$\Downarrow$$ $$HSO_4^{-} \rightarrow H^{+} + SO_4^{2-}$$
Example 2: Show the steps that a molecule of silicious acid will undergo to become a silicite ion, with the use of a series of chemical equations.
Solution: The silicious acid molecule has the chemical formula \(H_4SiO_3\), and according to the chart, can turn into \(SiO_3^{4-}\). With the possible removal of up to four hydrogen ions, silicious acid is a tetraprotic acid. We can write a series of four equations, where one hydrogen ion is removed from the silicious acid molecule in each one:
$$H_4SiO_3 \rightarrow H^{+} + H_3SiO_3^{-}$$ $$\Downarrow$$ $$H_3SiO_3^{-} \rightarrow H^{+} + H_2SiO_3^{2-}$$ $$\Downarrow$$ $$H_2SiO_3^{2-} \rightarrow H^{+} + HSiO_3^{3-}$$ $$\Downarrow$$ $$HSiO_3^{3-} \rightarrow H^{+} + SiO_3^{4-}$$
The next example involves a bit of stoichiometry, and thus is a quantitative problem, quite different from Examples 1 and 2.
Example 3: Phosphorous acid, \(H_3PO_4\), is a triprotic acid, releasing up to three hydrogen ions per molecule. Given this information, how many hydrogen ions can be formed by a sample of 2.5 moles of phosphorous acid molecules?
Solution: A mole refers to \(6.02 \times 10^{23}\) of something (this is called Avogadro's Number). Therefore, we can get rid of the mole, by converting from moles of phosphorous acid molecules to number of phosphorous acid molecules. We use this conversion:
$$2.5 \; mol \; H_3PO_4 \times \frac{6.02 \times 10^{23} \; H_3PO_4 \; molecules}{1 \; mol \; H_3PO_4} =$$ $$1.5 \times 10^{24} \; H_3PO_4 \; molecules$$
Now, in every molecule, there are three hydrogen atoms. Since phosphorous acid is a triprotic acid, there will be up to three hydrogen ions for every molecule. Thus our second and final conversion lets us change our units to hydrogen ions:
$$1.5 \times 10^{24} \; H_3PO_4 \; molecules \times \frac{3 \; H^{+} \; ions}{1 \; H_3PO_4 \; molecule} = 4.5 \times 10^{24} \; H^{+} \; ions$$
Polyprotic bases can accept more than one hydrogen ion. Often, these bases are hydroxides, ionic compounds with a hydroxide anion. Hydroxide ions prefer accepting hydrogen ions because the oxygen in hydroxide only has one bond, whereas it prefers to have two.
$$OH^{-} + H^{+} \rightarrow H_2O$$
A water molecule has two bonds on oxygen, making it more stable than the hydroxide ion. Polyprotic bases that are hydroxides will have more than one hydroxide ion. Some common hydroxides that are polyprotic bases are ionic compounds with the alkaline earth metals:
$$Mg(OH)_2$$ $$Ca(OH)_2$$ $$Ba(OH)_2$$ $$Zn(OH)_2$$ $$Cu(OH)_2$$ $$Al(OH)_3$$
Some transition metals such as zinc and copper, also commonly form hydroxides. These compounds all take at least two hydrogen ions from another source (such as an acid) to turn all the hydroxide ions into water.
Example 4: Calcium hydroxide, \(Ca(OH)_2\), is a common base. Add as many hydrogen ions as you need to turn all the hydroxide ions in one molecule into water. Give a chemical equation for each step.
Solution: There are two hydroxide ions per molecule, so calcium hydroxide is a diprotic base. Each molecule will require two hydrogen ions to turn all the hydroxide ions in a molecule of calcium hydroxide into water. Add the hydrogen ions one at a time:
$$Ca(OH)_2 + H^{+} \rightarrow Ca^{2+} + OH^{-} + H_2O$$ $$\Downarrow$$ $$Ca^{2+} + OH^{-} + H_2O + H^{+} \rightarrow Ca^{2+} + 2H_2O$$
Example 5: Aluminum hydroxide has three hydroxide ions per molecule and is thus a triprotic base. Let \(X\) be a polyatomic ion such that the following acid-base reaction occurs:
$$H_3X + Al(OH)_3 \rightarrow H_2O + AlX$$
Given that this equation is balanced and the charge of every reactant and product is \(0\), give two possibilities of what polyatomic ion \(X\) could be.
Solution: The key step is to identify the charge of the polyatomic ion. The compound \(AlX\) is our hint; it implies that \(Al^{3+}\) and \(X\) are in a 1:1 ratio in this chemical reaction. \(X\) is a polyatomic ion with charge \(-3\), because then \(AlX\) is neutral. As default for acid-base reactions, an ionic compound is one of the products, and it must be \(AlX\), and all ionic compounds are defined as neutrally charged.
The table of polyatomic anions that form from polyprotic acids releasing hydrogen ions gives two possibilities for the mystery anion: \(PO_3^{3-}\) and \(PO_4^{3-}\).
As a check, we can substitute each of these anions into the equation to ensure it is balanced:
$$H_3PO_4 + Al(OH)_3 \rightarrow H_2O + AlPO_4$$ $$H_3PO_3 + Al(OH)_3 \rightarrow H_2O + AlPO_3$$
A carbonate is an ionic compound with a carbonate anion, \(CO_3^{2-}\). Carbonates are slightly more tricky to analyze in acid-base chemistry than hydroxides, because every carbonate ion will accept up to two hydrogen ions. That procedure is shown in Example 6.
Example 6: Carbonate ions are diprotic bases, accepting up to two hydrogen ions per molecule:
$$CO_3^{2-} + H^{+} \rightarrow HCO_3^{-}$$
This ion is known as the bicarbonate ion and will be studied more closely later. The second part of the reaction is
$$HCO_3^{-} + H^{+} \rightarrow H_2CO_3$$
The reason why carbonate ions will generally accept two hydrogen ions per molecule is because carbonic acid decomposes into carbon dioxide and water. So, a condensed form of this reaction is
$$CO_3^{2-} + 2H^{+} \rightarrow CO_2 + H_2O$$
This explains why every acid-base reaction involving a carbonate produces carbon dioxide in addition to water and an ionic compound. Here is one more example involving a carbonate:
Example 7: Is \(Al_2(CO_3)_3\) diprotic, tetraprotic, etc.? Write the chemical equation that shows this compound accepting the corresponding number of hydrogen ions.
Solution: There are three carbonate ions per molecule, and each carbonate ion requires two hydrogen ions to be neutralized. So six hydrogen ions are necessary, and the base is hexaprotic. Here is the overall reaction:
$$Al_2(CO_3)_3 + 6H^{+} \rightarrow 2Al + 3CO_2 + 3H_2O$$
A bicarbonate is an ionic compound with a bicarbonate ion. The bicarbonate ion is \(HCO_3^{-}\). Notice what happens if you add a hydrogen ion:
$$HCO_3^{-} + H^{+} \rightarrow H_2CO_3 \rightarrow CO_2 + H_2O$$
As a result, carbonate and bicarbonate ions are closely related, but a bicarbonate ion only requires one hydrogen ion to neutralize. As a result, some bicarbonates are monoprotic.
Example 8: \(Mg(HCO_3)_2\) requires how many hydrogen ions to neutralize? Write out the reaction.
Solution: There are two bicarbonate ions per molecule, so we need two hydrogen ions. The reaction:
$$Mg(HCO_3)_2 + 2H^{+} \rightarrow Mg^{2+} + CO_2 + H_2O$$
Brackets denote a concentration of an ion or molecule. With that in mind, the pH of a solution is defined as
$$pH = -\log_{10}[H^{+}]$$
The pH of a solution is influenced by whether the molecules are monoprotic or polyprotic. A polyprotic acid or base will have more than one hydrogen or hydroxide ion per molecule. This means that the \([H^{+}]\) or \([OH^{-}]\) (depending on whether the solution is acidic or basic) will be higher than the molarity.
If a molecule of an acid can release up to \(n\) hydrogen ions, then
$$[H^{+}] = nM$$
\(M\) denotes the molarity. If the solute of the solution is a base that can accept up to \(n\) hydrogen ions per molecule, then
$$[OH^{-}] = nM$$
Trying to find the pH or pOH will not be any more difficult, once you know how many hydrogen ions can be removed/accepted from a molecule of the solute. Also, notice all of this only applies to strong acids and bases where the hydrogen/hydroxide ions entirely separate from the molecules of the acid/base.
Example 8: What is the pH of a \(.054\) molar aqueous solution of sulfuric acid?
Solution: The chemical formula of sulfuric acid is \(H_2SO_4\). After removing two hydrogen ions, it becomes one of the common polyatomic ions, sulfate, so it is a diprotic acid, and
$$[H^{+}] = 2m = 2 \cdot .054 \; M = .108 \; M$$
With the hydrogen ion concentration, we can find the pH:
$$pH = -\log_{10}(.108 \; M) = .97$$
Just as a side note, be aware that the capital M is used to denote molarity as well as label a magnitude with the molarity unit.
Example 9: What is the pH of a \(.0915\) molar solution of calcium hydroxide?
Solution: Calcium hydroxide's chemical formula is \(Ca(OH)_2\). As a result it is a diprotic base, requiring two hydrogen ions to neutralize each molecule. Therefore,
$$[OH^{-}] = 2m$$
With this formula, we can find the hydroxide ion concentration:
$$[OH^{-}] = 2 \cdot .0915 \; M = .183 \; M$$
Now we can find the pH with the formula
$$pH = 14 + \log_{10}[OH^{-}]$$
The calculations proceed:
$$pH = 14 + \log_{10}(.183 \; M) = 13.26$$
Polyprotic acids and bases have different properties than monoprotic acids and bases. More hydrogen ion transfer is required to neutralize an acid or a base, and the concentrations of key ions will be different than a monoprotic acidic/basic aqueous solution of the same molarity. Therefore it is important to know the difference between monoprotic and polyprotic.