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Unit Four

Day 35: Acid and Base: Relative Strength and Reactions

As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Chapter 14-6 through 14-8, and/or Chapter 15.8-15.14 in the Additional Reading Materials section.

D35.1 Polyprotic Acids

We can classify acids by the number of protons per molecule that they can donate in an acid-base reaction. Acids that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Examples are HCl, HNO3, and HCN

Even though it contains four hydrogen atoms, acetic acid 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. The dissociation of the first H+ always takes place to a greater extent than the dissociation of the second H+. For example, sulfuric acid ionizes in two steps:

H2SO4(aq) + H2O(l) ⇌ HSO4–(aq) + H3O+(aq)     Ka,1 > 102
HSO4–(aq) + H2O(l) ⇌ SO42-(aq) + H3O+(aq)     Ka,2 = 1.1 × 10-2

This stepwise ionization 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. For example, when carbonic acid loses one H+, it yields hydronium ions and bicarbonate ions in small quantities:

H2CO3(aq) + H2O(l) ⇌ HCO3–(aq) + H3O+(aq)     [latex]K_{\text{H}_2\text{CO}_3} = \dfrac{[\text{H}_3\text{O}^{+}][\text{HCO}_3^{\;-}]}{[\text{H}_2\text{CO}_3]} = 4.3\;\times\;10^{-7}\;\text{M}[/latex]

The bicarbonate ion can lose an H+ to form hydronium ions and carbonate ions in even smaller quantities:

HCO3–(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)     [latex]K_{\text{HCO}_3^{\;-}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{CO}_3^{\;2-}]}{[\text{HCO}_3^{\;-}]} = 4.7\;\times\;10^{-11}\;\text{M}[/latex]

[latex]K_{\text{H}_2\text{CO}_3}[/latex] is larger than [latex]K_{\text{HCO}_3^{\;-}}[/latex] by a factor of about 104, so H2CO3 is the dominant producer of H3O+ in the solution. This means that the concentrations of H3O+ and HCO3– are practically equal in a pure aqueous solution of H2CO3.

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 H3O+ 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.

Example 1

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 CO2 reacts with water to form carbonic acid, H2CO3. A saturated solution of CO2 contains 0.033 mol H2CO3 per liter of water. Calculate [H3O+], [HCO3–], and  [CO32-] in a saturated solution of CO2.

Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than HCO3–, so H2CO3 is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Determine the concentration of H3O+ and HCO3– produced by ionization of H2CO3, assuming that further reaction of HCO3– is negligible. (2) Determine the concentration of CO32- in a solution with the concentration of H3O+ and HCO3– determined in (1).

  1. Determine the concentrations of H3O+ and HCO3–. Assume that further reaction of HCO3− is negligible.
    [latex]\text{H}_2\text{CO}_3(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{HCO}_3^{\;\;-}(\text{aq})\;\;\;\;\;\;\;K_{\text{a}1} = 4.3\;\times\;10^{-7}\;\text{M}[/latex]

    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium constant ( M ). The second column has the header of “H subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.

    Substituting in the equilibrium concentrations gives us:

    [latex]K_{\text{H}_2\text{CO}_3} = \dfrac{[\text{H}_3\text{O}^{+}][\text{HCO}_3^{\;\;-}]}{[\text{H}_2\text{CO}_3]}  \;\;\;\;\;\;\;\ \dfrac{(x)(x)}{0.033\;-\;x} = 4.3\;\times\;10^{-7}[/latex]

    Solving for x gives:

    [latex]x = 1.2\;\times\;10^{-4}[/latex]

    Thus:

    [latex][\text{H}_2\text{CO}_3] = 0.033\;M[/latex]
    [latex][\text{H}_3\text{O}^{+}] = [\text{HCO}_3^{\;\;-}] = 1.2\;\times\;10^{-4}\;M[/latex]
  2. Determine the concentration of CO32- in a solution at equilibrium with [H3O+] and [HCO3–], both equal to 1.2 ×10−4 M. (Note that, except for [CO32−], all concentrations and the equilibrium constant are known so no ICE table is needed.)
    [latex]\text{HCO}_3^{\;\;-}(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{CO}_3^{\;\;2-}(\text{aq})\;\;K_{\text{a}2} = 4.7\;\times\;10^{-11}\;\text{M}[/latex]
    [latex]K_{\text{HCO}_3^{\;\;-}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{CO}_3^{\;\;2-}]}{[\text{HCO}_3^{\;\;-}]} \;\;\;\;\;\;\;\; \dfrac{(1.2\;\times\;10^{-4}\;\text{M})[\text{CO}_3^{\;\;2-}]}{1.2\;\times\;10^{-4}\;\text{M}} = 4.7\;\times\;10^{-11}\;\text{M}[/latex]
    [latex][\text{CO}_3^{\;\;2-}] = \dfrac{(4.7\;\times\;10^{-11})(1.2\;\times\;10^{-4})}{1.2\;\times\;10^{-4}}\;\text{M} = 4.7\;\times\;10^{-11}\;\text{M}[/latex]

To summarize: In part 1 of this example, we found that the H2CO3 in a 0.033-M solution ionizes slightly and at equilibrium [H2CO3] = 0.033 M; [latex][\text{H}_3\text{O}^{+}] = 1.2\;\times\;10^{-4}\;\text{M}[/latex]; and [latex][\text{HCO}_3^{-}] = 1.2\;\times\;10^{-4}\;\text{M}[/latex]. In part 2, we determined that [latex][\text{CO}_3^{\;\;2-}] = 4.7\;\times\;10^{-11}\;\text{M}[/latex].

The assumption in part 1 that further reaction of HCO3− is negligible is valid because [latex][\text{CO}_3^{\;\;2-}] = 4.7\;\times\;10^{-11}\;\text{M}[/latex] << [latex][\text{HCO}_3^{-}] = 1.2\;\times\;10^{-4}\;\text{M}[/latex].
Also, [latex][\text{H}_3\text{O}^{+}] = 1.2\;\times\;10^{-4}\;\text{M}[/latex]; >> 10−7 M so autoionization of water can be ignored.

Check Your Learning
The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate [H3O+], [HS−], and [S2−] in the solution:

[latex]\begin{array}{rcll} \text{H}_2\text{S}(\text{aq})\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{HS}^{-}(\text{aq}) & K_{\text{a}1} = 8.9\;\times\;10^{-8} \;\text{M}\\[0.5em] \text{HS}^{-}(\text{aq})\;+\;\text{H}_2\text{O}(l) &\rightleftharpoons& \text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{S}_2^{\;\;-}(\text{aq}) & K_{\text{a}2} = 1.0\;\times\;10^{-19}\;\text{M} \end{array}[/latex]

Answer:

[H2S] = 0.1 M; [H3O+] = [HS−] = 9.4 × 10−5 M; [S2−] = 1 × 10−19 M

We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).

A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

H3PO4(aq) + H2O(l) ⇌ H2PO4–(aq) + H3O+(aq)     Ka,1 = 7.2 × 10-3 M
H2PO4–(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq)     Ka,2 = 6.3 × 10-8 M
HPO42-(aq) + H2O(l) ⇌ PO43-(aq) + H3O+(aq)        Ka,3 = 4.6 × 10-13 M

Again, the differences in the ionization constants of these reactions tell us that the degree of ionization is significantly weaker in each successive step. This is a general characteristic of polyprotic acids. Here, because the successive ionization constants differ by a factor of 105 to 106, the calculations of equilibrium concentrations in a solution of H3PO4 can be broken down into a series of parts, similar to those for diprotic acids.

Polyprotic bases can accept more than one H+. The carbonate ion is an example of a diprotic base, because it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:

CO32-(aq) + H2O(l) ⇌ HCO3–(aq) + OH–(aq)
HCO3–(aq) + H2O(l) ⇌ H2CO3(aq) + OH–(aq)

D35.2 Acid-Base Reactions

Mixing a solution of an acid with a solution of a base results in an acid-base neutralization reaction that produces a salt and water. A general rule for acid-base reactions is that a stronger acid and a stronger base react to form a weaker conjugate base and a weaker conjugate acid. If the stronger acid and stronger base are on the left side of an equilibrium reaction, the reaction is product-favored; if the stronger adic and base are on the right side, the reaction is reactant-favored. Strengths of acids and bases are measured by the corresponding Ka and Kb values, which can be obtained from a table.

A strong acid and a strong base react to form a neutral solution (containing equal concentrations of hydronium and hydroxide ions) if we mix stoichiometrically equivalent quantities. For example:

HCl(aq) + NaOH(aq) ⇌ NaCl(aq) + H2O(l)

The salt formed, NaCl, consists of ions, Na+(aq) and Cl−(aq), that have negligible acid or base strength so this equilibrium heavily favors the product side and goes essentially to completion. If the mixture has an excess of one of the reactants, then the concentration of leftover acid (HCl) or base (NaOH) determines the pH of the final solution.

A weak acid and a strong base, react to form a salt that contains the conjugate base of the acid, which is usually a weak base. For example, the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

CH3COOH(aq) + NaOH(aq) ⇌ CH3COONa(aq) + H2O(l)

forms sodium acetate. The equilibrium of this reaction favors the product side, and the reaction can be approximated as going to complection. Mixing stoichiometrically equivalent amounts of reactants gives a solution containing Na+(aq), which  has no effect on the pH of the solution, and CH3COO–(aq), the conjugate base of acetic acid. CH3COO–reacts with water and increases the concentration of hydroxide:

CH3COO–(aq) + H2O(l) ⇌ CH3COOH(aq) + OH–(aq)

Because acetate ion is a weak base, pH > 7 after acetic acid reacts stoichiometrically with a strong base. The equilibrium constant for reaction of acetate ion with water is the ionization constant, Kb, for the acetate anion. (Some reference tables only report ionization constants for acids; Kb can be calculated from Kw and Ka of the conjugate acid—acetic acid in this case.) Generalizing this example, when a strong base reacts stoichiometrically with a weak acid, the solution that results is basic. Also, if a solid salt that is the product of a weak acid-strong base reaction dissolves in water, the solution is basic.

Exercise 1: Relationship between Ionization Constants of an Acid and Its Conjugate Base

Write an equation showing the relationship between the pKa of an acid (HA) and the pKb of its conjugate base (A–)?

Example 3

Equilibrium in a Solution of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with [CH3COO–] = 0.050 M and [OH−] = 2.5 × 10−6 M at equilibrium. (Ka for acetic acid is 1.8 × 10-5 M) The reaction is:

[latex]\text{CH}_3\text{COO}^{-}(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{CH}_3\text{COOH}(\text{aq})\;+\;\text{OH}^{-}(\text{aq})[/latex]

Solution
We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:

[latex]K_{\text{b}}\;(\text{for CH}_3\text{COO}^{-}) = \dfrac{K_{\text{w}}}{K_{\text{a}}\;(\text{for CH}_3\text{CO}_2\text{H})} = \dfrac{1.0\;\times\;10^{-14}\;\text{M}^2}{1.8\;\times\;10^{-5}\;\text{M}} = 5.6\;\times\;10^{-10}\;\text{M}[/latex]

Now find the missing concentration:

[latex]K_{\text{b}} = \dfrac{[\text{CH}_3\text{CO}_2\text{H}][\text{OH}^{-}]}{[\text{CH}_3\text{CO}_2^{\;\;-}]} = 5.6\;\times\;10^{-10}\;\text{M}[/latex]
[latex]= \dfrac{[\text{CH}_3\text{CO}_2\text{H}](2.5\;\times\;10^{-6}\;\text{M})}{(0.050\;\text{M})} = 5.6\;\times\;10^{-10}\;\text{M}[/latex]

Solving this equation we get [CH3CO2H] = 1.1 × 10−5 M.

Check Your Learning
Calculate the pH of a 0.083-M solution of CN−. Use 3.3 × 10−10 M as Ka for HCN. Note that CN− is a weak base so pH > 7.

Answer:

11.20

If your answer does not agree, did you convert pOH to pH or find [H3O+] from [OH−]?

A strong acid reacts with a weak base to produce a solution of a salt containing the conjugate acid of the weak base, which is usually a weak acid. For example, the reaction of the weak base ammonia with the strong acid HCl,

NH3(aq) + HCl(aq) ⇌ NH4Cl(aq)

forms ammonium chloride. The equilibrium of this reaction favors the product side, and the reaction can be approximated as going to complection. Mixing stoichiometrically equivalent amounts of reactants gives a solution that contains Cl−(aq), which is the conjugate base of a strong acid and has no effect on the pH of the solution, and NH4+(aq), the conjugate acid of ammonia. NH4+ reacts with water and increases the hydronium ion concentration:

NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

Because ammonium ion is a weak acid, pH < 7 after ammonia reacts stoichiometrically with a strong acid. The equilibrium constant for the reaction of NH4+ with water is the ionization constant, Ka, for the acid NH4+. Generalizing this example, when a weak base reacts stoichiometrically with a strong acid, the solution that results is acidic. Also, if a solid salt that is the product of a weak base-strong acid reaction dissolves in water, the solution is acidic.

Example 4

The pH of a Solution of a Weak Base and a Strong Acid
Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [C6H5NH3]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. Calculate the pH of a 0.233 M solution of aniline hydrochloride. (Kb for aniline, C6H5NH2, is 3.9 × 10−10 M)

[latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}(\text{aq})\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(\text{aq})\;+\;\text{C}_6\text{H}_5\text{NH}_2(\text{aq})[/latex]

Solution
C6H5NH3+ ion is the conjugate acid of a weak base. We can determine value of Ka for C6H5NH3+ ion from the value of Kb for aniline:

[latex]K_{\text{a}}\;(\text{for C}_6\text{H}_5\text{NH}_3^{+})\;\times\;K_{\text{b}}\;(\text{for C}_6\text{H}_5\text{NH}_2) = K_{\text{w}} = 1.0\;\times\;10^{-14}\;\text{M}^2[/latex]
[latex]K_{\text{a}}\;(\text{for C}_6\text{H}_5\text{NH}_3^{\;\;+}) = \dfrac{K_{\text{w}}}{K_{\text{b}}\;(\text{for C}_6\text{H}_5\text{NH}_2)} = \dfrac{1.0\;\times\;10^{-14}\;\text{M}^2}{3.9\;\times\;10^{-10}\;\text{M}} = 2.6\;\times\;10^{-5}\;\text{M}[/latex]

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:
Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

With these steps we find [H3O+] = 2.4 × 10−3 M and pH = 2.62.

Check Your Learning
(a) Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of C6H5NH3+ is 2.4 × 10−3 M and the pH is 2.62.

(b) Calculate the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3–. The Kb for ammonia is 1.8 × 10−5. Which is the stronger acid C6H5NH3+ or NH4+?

Answer:

Ka(for NH4+) = 5.6 × 10-10; [H3O+] = 7.5 × 10−6 M; C6H5NH3+ is the stronger acid because it has larger Ka.

To predict the pH of a solution resulting from the reaction of a weak acid and a weak base, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic; if Kb > Ka, the solution is basic.

Example 5

Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO3

(c) NH4Cl

(d) Na2HPO4

(e) NH4F

Solution
Consider each of the ions separately in terms of its effect on the pH of the solution:

(a) Neither the K+ cation nor the Br− anion affects the acidity of a solution. K+ does not react with water and therefore does not affect pH. Br− is the conjugate base of a strong acid (HBr) and therefore has negligible strength as a base in aqueous solution. The solution is neutral.

(b) The Na+ cation does not affect the pH of the solution. The HCO3– anion is amphiprotic, it could either behave as an acid or a base. The Ka of HCO3– is 4.7 × 10−11 M, the Kb of HCO3– (from Ka of H2CO3) is [latex]\frac{1.0\;\times\;10^{-14}\;\text{M}^2}{4.3\;\times\;10^{-7}\;\text{M}} = 2.3\;\times\;10^{-8}\;\text{M}[/latex]. Because Kb > Ka, HCO3–  is more basic than acidic; the solution is basic.

(c) The NH4+ ion is acidic and the Cl− ion is a base of negligible strength. The solution is acidic.

(d) The Na+ ion has no acid-base properties. HPO42- is amphiprotic, with Ka = 4.6 × 10−13 M and Kb = 2.8 × 10-7 M. Because Kb > Ka, the solution is basic.

(e) The NH4+ ion is an acid and the F− ion is a base, so we must directly compare the Ka and the Kb of the two ions. Ka of NH4+ is 5.6 × 10−10 M, which seems very small, but the Kb of F− is 1.5 × 10−11 M, even smaller. So the solution is acidic, because Ka > Kb.

Check Your Learning
Determine whether aqueous solutions of eacg salt is acidic, basic, or neutral:

(a) K2CO3

(b) CaCl2

(c) KH2PO4

(d) (NH4)2CO3

Answer:

(a) basic; (b) neutral; (c) acidic; (d) basic

Exercise 2: Acid-Base Reactions

Complete each reaction below and predict whether the reaction is product-favored or reactant-favored.   (Enter formulas without subscripts and superscripts, and put charges in parentheses.  For example, you would enter “HSO4–” as “HSO4(-)”.)

D35.3 Reaction Between Amphiprotic Species

Acid-base reactions can also occur between two amphiprotic species. For example, mixing a solution containing hydrogen sulfate ions (HSO4–) and a solution containing hydrogen carbonate ions (HCO3–) results in an acid-base reaction. However, if both reactants can act as either an acid or a base, which reactant is the acid and which is the base? In the example mixture, there are two possibilities:

possibility I:      HSO4–(aq) + HCO3–(aq) ⇌ SO42-(aq) + H2CO3(aq)
possibility II:     HSO4–(aq) + HCO3–(aq) ⇌ H2SO4(aq) + CO32-(aq)

Qualitatively, a product-favored acid-base reaction involves a stronger acid reacting with a stronger base to form a weaker acid and a weaker base. Acid strengths are measured by Ka values and base strengths by Kb values. In possibility I the acids are HSO4− (Ka = 1.1 × 10-2 M) and H2CO3 (Ka = 4.3 × 10-7 M) and the bases are HCO3− (Kb = 2.3 × 10-8 M) and SO42− (Kb = 9.1 × 10-13 M). The stronger acid and the stronger base are on the left side of the equation so this reaction is product-favored. The second reaction is reactant-favored because it produces H2SO4, a strong acid, and CO32-, a weak base with the relatively large Kb = 2.1 × 10−4 M (significantly larger than the very small Kb  for HSO4–).

Quantitatively, we can make use of the ionization constants to determine which reaction occurs. In possibility I,

HSO4–(aq) + H2O(l) ⇌ SO42-(aq) + H3O+(aq)      [latex]K_1 = K_{\text{a, HSO}_4^-} = 1.1\;\times\;10^{-2}\;\text{M}[/latex]
HCO3–(aq) + H3O+(aq) ⇌ H2CO3(aq) + H2O(l)      [latex]K_2 = \dfrac{1}{K_{\text{a, H}_2\text{CO}_3}} = \dfrac{1}{4.3\;\times\;10^{-7}\;\text{M}}[/latex]

The sum of these two equilibria, the reaction HSO4–(aq) + HCO3–(aq) ⇌ SO42-(aq) + H2CO3(aq), has a total equilibrium constant of:

Ktotal, possibility I = K1 × K2 = (1.1 × 10-2 M)/(4.3 × 10-7 M) = 2.6 × 104

Clearly possibility I is product-favored because the equilibrium constant is much greater than 1.

In possibility II,

HSO4–(aq) + H3O+(aq) ⇌ H2SO4(aq) + H2O(l)         [latex]K_1 = \dfrac{1}{K_{\text{a, H}_2\text{SO}_4}} ≤ \dfrac{1}{20\;\text{M}}[/latex]
HCO3–(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)         [latex]K_2 = K_{\text{a, HCO}_3^-} = 4.7\;\times\;10^{-11}\;\text{M}[/latex]

(Ka for H2SO4 is too large to measure in aqueous solution but is greater than Ka for HNO3, which is ≅ 20 M, so the value 20 M is a minimum for Ka for H2SO4.) The sum of these two equilibria, the reaction HSO4–(aq) + HCO3–(aq) ⇌ H2SO4(aq) + CO32-(aq), has a total equilibrium constant of:

Ktotal, II = K1 × K2 = (4.7 × 10-11 M)/(20 M) ≤ 2 × 10−12

Possibility I is product-favored but possibility II is not. Therefore, HSO4– acts as an acid and HCO3– acts as a base.

D35.4 Amino Acids

Amino acids are also amphiprotic. Each amino acid molecule contains a carboxyl group and an amino group. We have seen that carboxylic acids are moderately acidic, many with pKa of ~5. We have also seen that organic amines are somewhat basic, many with pKb of about 4 to 5. This combination creates an interesting situation, where an acid-base reaction is possible within a single amino acid molecule:realalanine.gif
The carboxylic acid, with pKa approximately 5, is a stronger acid than the protonated amine group, with pKa = 14 – pKb(amine) ≅ 9. The amine group (pKb about 4 to 5) is a stronger base than the carboxylate ion (pKb = 14 − 5 = 9). The stronger acid and stronger base are on the left side of the equilibrium so it is product-favored and the amino acid molecule contains both a positively charged and a negatively charged group. At the pH of a typical living organism, the amino acid is a zwitterion (German for “double ion”). A zwitterion is a species with no overall electrical charge but with separate parts that are positively and negatively charged.

The formation of a zwitterion is analogous to the acid-base reaction between methylamine (Kb = 4.4 × 10-4) and acetic acid (Ka = 1.8 × 10-5):

CH3NH2(aq) + CH3COOH(aq) ⇌ CH3NH3+(aq) + CH3COO–(aq)

where the equilibrium favors products because:

[latex]K_{\text{total}} = \dfrac{1.8\;\times\;10^{-5}\;\text{M}}{2.3\;\times\;10^{-11}\;\text{M}} = 7.8\;\times\;10^{5}[/latex]

Increasing the pH of an amino acid solution by adding hydroxide ions can remove the hydrogen ion from the -NH3+ group:

zwittbase.gif

Decreasing the pH by adding strong acid to an amino acid solution protonates the -COO– part of the zwitterion.
zwittacid.gif

Podia Question

Write a clear, concise explanation in scientifically appropriate language for each of these correct statements.

1. When a polyprotic acid donates a hydrogen ion, the species that remains is usually a much weaker acid than was the original polyprotic acid.

2. When trichloroacetic acid reacts with hydrogen carbonate ion the equilibrium is significantly more product-favored than when acetic acid reacts with hydrogen carbonate ion.

3. When a strong base reacts with a weak acid in stoichiometrically equivalent quantity, the pH of the solution is above 7.

Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.

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