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By the end of this section, you will be able to:
  • Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton

We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO 3 , and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids . Their reactions with water are:

HCl ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + Cl ( a q ) HNO 3 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + NO 3 ( a q ) HCN ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + CN ( a q )

Even though it contains four hydrogen atoms, acetic acid, CH 3 CO 2 H, is also monoprotic because only the hydrogen atom from the carboxyl group (COOH) reacts with bases:

This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).

Similarly, monoprotic bases are bases that will accept a single proton.

Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:

First ionization: H 2 SO 4 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + HSO 4 ( a q ) K a 1 = more than 10 2 ; complete dissociation Second ionization: HSO 4 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + SO 4 2− ( a q ) K a 2 = 1.2 × 10 −2

This stepwise ionization    process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, H 2 CO 3 , is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

First ionization: H 2 CO 3 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + HCO 3 ( a q ) K H 2 CO 3 = [ H 3 O + ] [ HCO 3 ] [ H 2 CO 3 ] = 4.3 × 10 −7

The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.

Second ionization: HCO 3 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + CO 3 2− ( a q ) K HCO 3 = [ H 3 O + ] [ CO 3 2− ] [ HCO 3 ] = 5.6 × 10 −11

K H 2 CO 3 is larger than K HCO 3 by a factor of 10 4 , so H 2 CO 3 is the dominant producer of hydronium ion in the solution. This means that little of the HCO 3 formed by the ionization of H 2 CO 3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H 3 O + and HCO 3 are practically equal in a pure aqueous solution of H 2 CO 3 .

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H 3 O + and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.

Ionization of a diprotic acid

When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO 2 reacts with water to form carbonic acid, H 2 CO 3 . What are [ H 3 O + ] , [ HCO 3 ] , and [ CO 3 2− ] in a saturated solution of CO 2 with an initial [H 2 CO 3 ] = 0.033 M ?

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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