Question

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 man suffering from indigestion can be considered to be 200.mL of a 0.012 M HCl solution. What mass of NaHCO3 would he need to ingest to neutralize this much HCl ? Round your answer to 2 significant digits.

Answer 1

Answer:

$\boxed{\text{0.20 g}}$

Explanation:

We know we will need a balanced chemical equation with molar masses, volumes, and concentrations, so, let's gather all the information in one place.

M_r:                       84.01

                 HCl + NaHCO₃ ⟶ NaCl + H₂O + CO₂

V/mL:        200.

c/mol·L⁻¹:  0.012

(a) Moles of HCl

$\text{Moles of HCl} =\text{0.200 L HCl} \times \dfrac{\text{0.012 mol HCl}}{\text{1 L HCl}}\\\\=\text{0.0024 mol HCl}$

(b) Moles of NaHCO₃

The molar ratio is 1 mol NaHCO₃ = 1 mol HCl

$\text{Moles of NaHCO _{3} }= \text{0.0024 mol HCl} \times \dfrac{\text{1 mol {NaHCO _{3} }}}{ \text{1 mol HCl}}\\\\= \text{0.0024 mol NaHCO _{3} }$

(c) Mass of NaHCO₃

$\text{Mass of NaHCO _{3} }= \text{0.0024 mol NaHCO _{3} } \times \dfrac{\text{84.01 g {NaHCO _{3} }}}{ \text{1 mol NaHCO _{3} }}\\\\= \textbf{0.20 g NaHCO _{3} }\\\\\text{The man would need to ingest }\boxed{\textbf{0.20 g}} \text{ of NaHCO _{3} }.$