Question

How many Tums tablets containing 750 mg calcium carbonate would be needed to raise the pH of the stomach to 1.5 if the stomach contains 2.0 L of gastric juice with a pH of 1.2. How many to raise the pH to 3.7? Show steps.

Answer 1

stomach contains 2.0 L of gastric juice with a pH of 1.2

pH = -log [H⁺]

[H⁺]= 10^-pH

[H⁺]= 10^-1.2

[H⁺]= 0.0631 M

When the pH = 1.5 the [H⁺] should be

pH = -log [H⁺]

[H⁺]= 10^-pH

[H⁺]= 10^-1.5

[H⁺]= 0.0316 M

Difference is 0.0631 M - 0.0316 M = 0.0316 in 2 L it will be 0.0316 x 2 = 0.063 moles

0.063 moles of calcium carbonate is

MW of calcium carbonate is 100g/mol 0.063 moles will be = 0.063 x 100 = 6.3 g

1 Tablet has 750 mg

6.3 g is 8.4 tablets

For pH 3.7 it will be

pH = -log [H⁺]

[H⁺]= 10^-pH

[H⁺]= 10^-3.7

[H⁺]= 1.99 x 10^-4 M

Difference is 0.0631 M - 1.99 x 10^-4 M = 0.0629 in 2 L it will be 0.0316 x 2 = 0.1258 moles

MW of calcium carbonate is 100g/mol 0.1258 moles will be = 0.1258 x 100 = 12.58 g

1 Tablet has 750 mg

12.58 g is 16.77 tablets