Question 21.2: Acid reflux (sometimes called heartburn, though the heart is...
Acid reflux (sometimes called heartburn, though the heart is not involved) affects many people. It results from acid in the stomach leaking into the esophagus and causing discomfort. Stomach acid is primarily an aqueous solution of HCl. (a) Calculate the molarity of hydrochloric acid in gastric juice that has a pH of 0.80. (b) One treatment for the symptoms of acid reflux is to take an antacid tablet. What volume of gastric juice can be neutralized by a 750-mg tablet of calcium carbonate (a typical size for an overthe- counter antacid)?
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Collect and Organize We are given the pH of a solution and are asked to calculate the concentration of HCl that corresponds to that pH. As we saw in Chapter 15, pH =- \log[H_{3}O^{+}]. Hydrochloric acid is a strong acid and ionizes completely to H_{3}O^{+} and Cl^{-} in water:
HCl(aq) + H_{2}O(\ell) → H_{3}O^{+}(aq) + Cl^{-}(aq)
We are also asked to calculate the volume of HCl solution that can be neutralized by a 750-mg tablet of calcium carbonate, CaCO_{3}. We need to write a balanced chemical equation for the neutralization reaction.
Analyze The equation describing the ionization of hydrochloric acid indicates that 1 mole of H_{3}O^{+} ions is formed for every mole of HCl present. The pH of gastric juice falls between 1 and 0, so [H_{3}O^{+}] will be between 10^{-1} (= 0.1) M and 10^{0} (= 1) M.
The neutralization reaction is
CaCO_{3}(s) + 2 H_{3}O^{+}(aq) → Ca^{2+}(aq) + CO_{2}(g) + 3 H_{2}O(\ell)
This equation indicates that 2 moles of H_{3}O^{+} are consumed for every mole of CaCO_{3}.
We are told that the tablet size is typical of an antacid tablet, so common sense leads us to predict that the volume of acid this tablet can neutralize will not be excessively large (greater than 1 L) or small (less than 10 mL): too large a tablet would be a waste of antacid, and too small a tablet would not relieve the symptoms.
Solve
a. Substitution into Equation 15.10 gives
pH =- log[H_{3}O^{+}] (15.10)
pH =- log[H_{3}O^{+}] = 0.80
We take the antilog of both sides to solve for [H_{3}O^{+}]:
[H_{3}O^{+}]=10^{-0.80} = 0.16 M H_{3}O^{+}
Therefore the concentration of HCl is
0.16 M H_{3}O^{+} \times \frac{1 mol HCl}{1 mol H_{3}O^{+}}=0.16 M HCl
b. First we calculate the number of moles of CaCO_{3} present in 750 mg:
0.750 \sout{g CaCO_{3}} \times \frac{1 mol CaCO_{3}}{100.09 \sout{g CaCO_{3}}}=7.49 \times 10^{-3} mol CaCO_{3}
Next we use the stoichiometry of the neutralization reaction to calculate the volume of 0.16 M HCl this quantity of CaCO_{3} can neutralize:
7.49 \times 10^{-3} \sout{mol CaCO_{3}} \times \frac{2 \sout{mol H_{3}O^{+}}}{1 \sout{mol CaCO_{3}}} \times \frac{1 L}{0.16 \sout{mol H_{3}O^{+}}}
=9.36 \times 10^{-2} L = 94 mL of 0.16 M HCl
Think About It A concentration of 0.16 M seems reasonable because it is indeed within the range of values predicted for a solution with pH < 1. The volume of 0.16 M acid that a 750-mg tablet of CaCO_{3} can neutralize is also reasonable; 94 mL represents about 3 ounces of gastric juice.