The daily output of stomach acid (gastric juice) is 1000 to 2000 mL. Prior to a

Question

The daily output of stomach acid (gastric juice) is 1000 to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1.38.

A.) What is the [H3O+] of stomach acid?

B.) One chewable tablet of the antacid Maalox contains 600. mg of CaCO3. Enter the neutralization equation.

C.) Given that one chewable tablet of the antacid Maalox contains 600. mg of CaCO3, calculate the milliliters of stomach acid neutralized by two tablets of Maalox.

D.) The antacid milk of magnesia contains 400. mg of Mg(OH)2 per teaspoon. Enter the neutralization equation.

E.) Given that the antacid milk of magnesia contains 400. mg of Mg(OH)2 per teaspoon, calculate the number of milliliters of stomach acid that are neutralized by 1 tablespoon of milk of magnesia. (1 tablespoon = 3 teaspoons.)

Explanation / Answer

Ans. #A. [H3O+] = antilog (-pH) = antilog (-1.38) = 0.0417 M

#B. Balanced reaction: CaCO3(s) + 2 HCl(aq) ----> CaCl2(aq) + H2O(l) + CO2(g)

#C. Total mass of CaCO3 in 2 chewable tablets = 2 x Mass of CaCO3 in one tablet

                                                = 2 x 600.0 mg

                                                = 1200.0 mg

                                                = 1.200 g

Moles of CaCO3 = Mass/ Molar mass

                                                = 1.200 g / (100.0872 g/mol)

                                                = 0.012 mol

According to the stoichiometry of balanced reaction, 1 moll CaCO3 neutralizes 2 mol HCl.

Hence, moles of HCl neutralized = 2 x total moles of CaCO3

                                                = 2 x 0.012 mol

                                                = 0.024 mol

Now,

Required volume of stomach acid = Moles of HCl / Molarity of [H3O+] in gastric juice

                                                = 0.024 mol / (0.0417 M)                            ; [1 M = 1 mol/ L]

                                                = 0.024 mol / (0.0417 mol/ L)

                                                = 0.575539 L

                                                = 575.54 mL

#D. Balanced reaction: Mg(OH)2 + HCl(aq) -----> MgCl2(aq) + 2 H2O(l)

#E. Total amount of Mg(OH)2 = (400.0 mg / teaspoon) x 1 tablespoon

                                                = (400.0 mg / teaspoon) x 3 teaspoon

                                                = 1200.0 mg

                                                = 1.200 g

Moles of Mg(OH)2 = Mass/ Molar mass

                                                = 1.200 g / (58.31968 g/mol)

                                                = 0.0206 mol

According to the stoichiometry of balanced reaction, 1 moll Mg(OH)2 neutralizes 2 mol HCl.

Hence, moles of HCl neutralized = 2 x total moles of Mg(OH)2

                                                = 2 x 0.0206 mol

                                                = 0.0412 mol

Now,

Required volume of stomach acid = Moles of HCl / Molarity of [H3O+] in gastric juice

                                                = 0.0412 mol / (0.0417 M)                          ; [1 M = 1 mol/ L]

                                                = 0.0412 mol / (0.0417 mol/ L)

                                                = 0.988009 L

                                                = 988.01 mL

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