Stomach Antacids

Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO₃. The reaction,[latex]{\text{CaCO}}_{3}\left(s\right)+2\text{HCl}\left(aq\right)\rightleftharpoons {\text{aCl}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)+{\text{CO}}_{2}\left(g\right)[/latex]
not only neutralizes stomach acid, it also produces CO₂(g), which may result in a satisfying belch.

Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)₂. It works according to the reaction:[latex]\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{2+}\left(aq\right)+2{\text{OH}}^{\text{-}}\left(aq\right)[/latex]

The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :[latex]{\text{H}}_{3}{\text{O}}^{\text{+}}+{\text{OH}}^{\text{-}}\rightleftharpoons 2{\text{H}}_{2}\text{O}\left(l\right)[/latex]
This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.

Several antacids have aluminum hydroxide, Al(OH)₃, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.

Culinary Aspects of Chemistry

Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO₃ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 1). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.

Figure 1. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.

Example 1

The pH of a Solution of a Salt of a Weak Base and a Strong Acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\text{Cl}[/latex], a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?

[latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)[/latex]

Solution

The new step in this example is to determine K_a for the [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] ion. The [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] ion is the conjugate acid of a weak base. The value of K_a for this acid is not listed in Ionization Constants of Weak Acids, but we can determine it from the value of K_b for aniline, C₆H₅NH₂, which is given as 4.6 [latex]\times[/latex] 10⁻¹⁰ (Relative Strengths of Acids and Bases and Ionization Constants of Weak Bases):

[latex]{K}_{\text{a}}\left(\text{for}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\times {K}_{\text{b}}\left(\text{for}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)={K}_{\text{w}}=1.0\times {10}^{-14}[/latex]
[latex]{K}_{\text{a}}\left(\text{for}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)=\frac{{K}_{\text{w}}}{{K}_{\text{b}}\left(\text{for}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)}=\frac{1.0\times {10}^{-14}}{4.6\times {10}^{-10}}=2.2\times {10}^{-5}[/latex]

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex], and the pH:

With these steps we find [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 2.3 [latex]\times[/latex] 10⁻³M and pH = 2.64

Check Your Learning

(a) Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] is 2.3 [latex]\times[/latex] 10⁻³ and the pH is 2.64.

(b) What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH₄NO₃, a salt composed of the ions [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] and [latex]{\text{NO}}_{3}{}^{\text{-}}[/latex]. Use the data in Table 2 on Relative Strengths of Acids and Bases to determine K_b for the ammonium ion. Which is the stronger acid [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] or [latex]{\text{NH}}_{4}{}^{\text{+}}?[/latex]

Answer: (a) [latex]{K}_{a}\left(\text{for}{\text{NH}}_{4}{}^{\text{+}}\right)=5.6\times {10}^{-10}[/latex], [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] = 7.5 [latex]\times[/latex] 10⁻⁶M; (b) [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}[/latex] is the stronger acid.

Example 2

Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

Determine the acetic acid concentration in a solution with [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right]=0.050M[/latex] and [OH⁻] = 2.5 [latex]\times[/latex] 10⁻⁶M at equilibrium. The reaction is:[latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)[/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 K_b as follows:[latex]{K}_{\text{b}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right)=\frac{{K}_{\text{w}}}{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}[/latex]

Now find the missing concentration:

[latex]{K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right]}=5.6\times {10}^{-10}[/latex]
[latex]=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left(2.5\times {10}^{-6}\right)}{\left(0.050\right)}=5.6\times {10}^{-10}[/latex]

Solving this equation we get [CH₃CO₂H] = 1.1 [latex]\times[/latex] 10⁻⁵M.

Check Your Learning

What is the pH of a 0.083-M solution of CN⁻? Use 4.0 [latex]\times[/latex] 10⁻¹⁰ as K_a for HCN. Hint: We will probably need to convert pOH to pH or find [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] using [OH⁻] in the final stages of this problem.

Answer: 11.16

Example 3

Determining the Acidic or Basic Nature of Salts

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

1. KBr
2. NaHCO₃
3. NH₄Cl
4. Na₂HPO₄
5. NH₄F

Solution

Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:

1. The K⁺ cation and the Br⁻ anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.
2. The Na⁺ cation is a spectator, and will not affect the pH of the solution; while the [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] anion is amphiprotic, it could either behave as an acid or a base. The K_a of [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] is 4.7 [latex]\times[/latex] 10⁻¹¹, so the K_b of its conjugate base is [latex]\frac{1.0\times {10}^{-14}}{4.7\times {10}^{-11}}=2.1\times {10}^{-4}[/latex].
   Since K_b >> K_a, the solution is basic.
3. The [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] ion is acidic and the Cl⁻ ion is a spectator. The solution will be acidic.
4. The Na⁺ ion is a spectator, while the [latex]{\text{HPO}}_{4}{}^{\text{-}}[/latex] ion is amphiprotic, with a K_a of 3.6 [latex]\times[/latex] 10⁻¹³
   so that the K_b of its conjugate base is [latex]\frac{1.0\times {10}^{-14}}{3.6\times {10}^{-13}}=2.8\times {10}^{-2}[/latex]. Because K_b >> K_a, the solution is basic.
5. The [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] ion is listed as being acidic, and the F⁻ ion is listed as a base, so we must directly compare the K_a and the K_b of the two ions. K_a of [latex]{\text{NH}}_{4}{}^{\text{+}}[/latex] is 5.6 [latex]\times[/latex] 10⁻¹⁰, which seems very small, yet the K_b of F⁻ is 1.4 [latex]\times[/latex] 10⁻¹¹, so the solution is acidic, since K_a > K_b.

Check Your Learning

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

1. K₂CO₃
2. CaCl₂
3. KH₂PO₄
4. (NH₄)₂CO₃
5. AlBr₃

Answer: (1) basic; (2) neutral; (3) basic; (4) basic; (5) acidic

Example 4

Hydrolysis of [Al(H₂O)₆]³⁺

Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [latex]{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}[/latex] in solution.

Solution

In spite of the unusual appearance of the acid, this is a typical acid ionization problem.

1. Determine the direction of change. The equation for the reaction and K_a are:[latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\left(aq\right){K}_{\text{a}}=1.4\times {10}^{-5}[/latex]The reaction shifts to the right to reach equilibrium.
2. Determine x and equilibrium concentrations. Use the table:
3. Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:[latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\right]}{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\right]}[/latex][latex]=\frac{\left(x\right)\left(x\right)}{0.10-x}=1.4\times {10}^{-5}[/latex]Solving this equation gives:[latex]x=1.2\times {10}^{-3}M[/latex]From this we find:[latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.2\times {10}^{-3}M[/latex][latex]\text{pH}=\text{-log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=2.92\left(\text{an acidic solution}\right)[/latex]
4. Check the work. The arithmetic checks; when 1.2 [latex]\times[/latex] 10⁻³M is substituted for x, the result = K_a.

Check Your Learning

What is [latex]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{2+}\right][/latex] in a 0.15-M solution of Al(NO₃)₃ that contains enough of the strong acid HNO₃ to bring [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] to 0.10 M?

Answer: 2.1 [latex]\times[/latex] 10⁻⁵M