How many moles of stomach acid would be neutralized by one tablet of Tums Ultra

Question

How many moles of stomach acid would be neutralized by one tablet of Tums Ultra 1000 that contains 1000 mg of calcium carbonate?

Assuming the volume of the stomach to be 1.0L, what will be the pH change of the stomach acid resulting from the ingestion of one Tums Ultra tablet that contains 1000 mg of calcium carbonate?

Please if you could explain how to do it.... Thank you Chemistry God or Goddess!!!!

a) How many moles of stomach acid would be neutralized by one tablet of Tums Ultra 1000 that contains 1000 mg of calcium carbonate? CaC03 (aq) + 2 H30+ (aq) Ca2+ (aq) + CO2(g) + 3 H20(1) b) Assuming the volume of the stomach to be 1.0L, what will be the pH change of the stomach acid resulting from the ingestion of one Tums Ultra tablet that contains 1000 mg of calcium carbonate.

Explanation / Answer

(a) According to given balanced chemical reaction equation

CaCO3(aq) + 2 H3O+(aq) --> Ca2+(aq) + CO2(g) + 3H2O(l)

the moles ratio of CaCO3: H3O+ is 1:2, this represents 1 mole CaCO3 requires 2 mole H+ (H+ = H3O+).

Now we will calculate moles present in 1000 mg Tums Ultra Tablet contains CaCO3

moles CaCO3 = mass of CaCO3 / molar mass of CaCO3

1 gram = 1000 mg ==>

moles CaCO3 = (1 g / 100.0869 g/mol) = 0.00999131754 moles

from above ratio of CaCO3: H3O+ = 1:2, now we have 0.00999131754 moles of CaCO3 therefore we need 2 x 0.00999131754 moles = 0.01998263509 moles of stomach acid ~ = 1.998 x 10-2 moles

moles of stomach acid would be neutralized by one tablet of Tums Ultra 1000 that contains 1000 mg of calcium carbonate = 1.998 x 10-2 moles

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(b) It is known that the main acidic component of gastric acid is hydrochloric acid and pH of the solution is about 1.5.

pH = -log[H+]

[H+] = 10-pH = 10-1.5 = 0.032 M

0.032 M means 0.032 moles / liter, we have already 1.0 L stomach acid,

hence moles present in 1.0 L stomach is 0.032 moles. according to the ratio of CaCO3: H3O+ = 1:2

moles CaCO3 required = 0.032 moles / 2 = 0.016 moles, but we do not have that much moles (we have only 0.00999131754 moles). Therefore we have excess acid moles, let us find out how many acid moles remained after reacting with 0.00999131754 moles CaCO3

moles H3O+ remains = 0.032 moles - 0.01998263509 moles [answer from (a)] = 0.0120173649 moles

pH = -log[H3O+] = -log[0.0120173649]

pH = 1.92

Stomach acid initial pH = 1.5, now it is increased to 1.92, so pH change = final - initial = 1.92 - 1.5 = 0.42

The pH change of the stomach acid resulting from the ingestion of one Tums Ultra tablet that contains 1000 mg of calcium carbonate = 0.42

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