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

Question

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)?

Accepted Answer

Collect and Organize We are given the pH of a solution and are asked to calculate the concentration of [latex]HCl[/latex] that corresponds to that pH. As we saw in Chapter 15, [latex]pH =- \log[H_{3}O^{+}][/latex]. Hydrochloric acid is a strong acid and ionizes completely to [latex]H_{3}O^{+}[/latex] and [latex]Cl^{-}[/latex] in water:

[latex]HCl(aq) + H_{2}O(\ell) → H_{3}O^{+}(aq) + Cl^{-}(aq)[/latex]

We are also asked to calculate the volume of [latex]HCl[/latex] solution that can be neutralized by a 750-mg tablet of calcium carbonate, [latex]CaCO_{3}[/latex]. 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 [latex]H_{3}O^{+}[/latex] ions is formed for every mole of [latex]HCl[/latex] present. The pH of gastric juice falls between 1 and 0, so [latex][H_{3}O^{+}][/latex] will be between [latex]10^{-1} (= 0.1) M[/latex] and [latex]10^{0} (= 1) M[/latex].

The neutralization reaction is

[latex]CaCO_{3}(s) + 2 H_{3}O^{+}(aq) → Ca^{2+}(aq) + CO_{2}(g) + 3 H_{2}O(\ell) [/latex]

This equation indicates that 2 moles of [latex]H_{3}O^{+}[/latex] are consumed for every mole of [latex]CaCO_{3}[/latex].
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

[latex]pH =- log[H_{3}O^{+}] [/latex] (15.10)

[latex]pH =- log[H_{3}O^{+}] = 0.80 [/latex]

We take the antilog of both sides to solve for [latex][H_{3}O^{+}][/latex]:

[latex][H_{3}O^{+}]=10^{-0.80} = 0.16 M H_{3}O^{+}[/latex]

Therefore the concentration of HCl is

[latex]0.16 M H_{3}O^{+} imes \frac{1 mol HCl}{1 mol H_{3}O^{+}}=0.16 M HCl[/latex]

b. First we calculate the number of moles of [latex]CaCO_{3}[/latex] present in 750 mg:

[latex]0.750 \sout{g CaCO_{3}} imes \frac{1 mol CaCO_{3}}{100.09 \sout{g CaCO_{3}}}=7.49 imes 10^{-3} mol CaCO_{3}[/latex]

Next we use the stoichiometry of the neutralization reaction to calculate the volume of 0.16 M HCl this quantity of [latex]CaCO_{3}[/latex] can neutralize:

[latex]7.49 imes 10^{-3} \sout{mol CaCO_{3}} imes \frac{2 \sout{mol H_{3}O^{+}}}{1 \sout{mol CaCO_{3}}} imes \frac{1 L}{0.16 \sout{mol H_{3}O^{+}}}[/latex]

[latex]=9.36 imes 10^{-2} L = 94 mL[/latex] 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 [latex]pH < 1[/latex]. The volume of 0.16 M acid that a 750-mg tablet of [latex]CaCO_{3}[/latex] can neutralize is also reasonable; 94 mL represents about 3 ounces of gastric juice.

Question 21.2: Acid reflux (sometimes called heartburn, though the heart is...

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