Sodium hydrogen carbonate , also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid ind

Question

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Sodium hydrogen carbonate , also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid ind

igestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved neutralizes excess through this reaction: (aq) (aq) (aq) (l) (g) The gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be of a M solution. What mass of would she need to ingest to neutralize this much ? Round your answer to significant digits.

Chemistry

1 answer:

Yanka [14] · Answer 1 · 2022-07-25T08:05:27+03:00

Answer:

The mass of NaHCO3 that she would need to ingest to neutralize this much HCl is 1.512g

Explanation:

The question can be better presented as follows:

Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid HCl, which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g)

The CO2 gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 200.mL of a 0.089M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl? Be sure your answer has the correct number of significant digits.

SOLUTION

Given:

Volume of HCl = 200ml = 0.200L

Molarity of the HCl =>0.089M

The number of mole of HCl can be calculated using C=n/V

Therefore n = CxV = 0.089 × 0.2

=0.018 moles

Equation for the reaction:

HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g)

From the equation, 1mole of HCl required 1mole of NaHCO3 for neutralization; therefore 0.018 moles of HCl will require 0.018 moles of NaHCO3.

But molar mass of NaHCO3 = 23+1+12+(16×3) = 84

Finally using number of moles = mass/molar mass