Question

Suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 50. ml. of a 0.048 M HCl solution. What mass of NaHCO3 would

he need to ingest to neutralize this much HCI? Round your answer to 2 significant digits.
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Answer 1

To neutralize the HCl in the stomach fluid, 0.048 M HCl solution, the individual would need to ingest 0.20 grams of NaHCO3.

Step-by-step explanation:

To neutralize the 0.048 M HCl solution, we can use the balanced chemical equation between HCl and NaHCO3:

2NaHCO3 + HCl → Na2CO3 + 2H2O + CO2

From the balanced equation, we can see that 2 moles of NaHCO3 react with 1 mole of HCl. To find the mass of NaHCO3 needed, we first calculate the number of moles of HCl in the 50 mL solution:

Moles of HCl = concentration (M) x volume (L)

Moles of HCl = 0.048 M x 0.050 L = 0.0024 mol HCl

Therefore, the mass of NaHCO3 needed is:

Mass of NaHCO3 = moles of HCl x molar mass of NaHCO3

Using the molar mass of NaHCO3 (84.01 g/mol),

Mass of NaHCO3 = 0.0024 mol x 84.01 g/mol = 0.20 g

Rounded to 2 significant digits, the mass of NaHCO3 needed is 0.20 g.

Learn more about Neutralizing acids and bases