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 0.035 M HCl solution. What mass of NaHCO3 would he need to ingest to neutralize this much HCl ? Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The mass of NaHCO₃ that the man would need to ingest to neutralize this much HCl is 0.059 g.

Explanation:

1) Chemical equation (given):

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

Both sides have the same number of atoms of each element:

Element       Let side      Right side

H                  1 + 1 = 2       1 × 2 = 2

Cl                 1                   1

Na                1                   1

C                  1                   1

O                 3                   1 + 2 = 3

Hence, the equation is balanced.

2. Mole ratios:

• 1 mol HCl  : 1 mol NaHCO₃ : 1 mol NaCl : 1 mol H₂O : 1 mol CO₂

2) Determine the number of moles of HCl in solution:

• M = n / V (liters) ⇒ n = M × V (liters)
• V (liters) = 200.mL (1 liter / 1000 mL) = 0.200 liter
• n = 0.035 M  × 0.200 liter = 0.00070 mol of HCl

3) Set a proportion with the stoichiometric ratio and the actual ratio:

• 1 mol NaHCO₃ / 1 mol HCl = x / 00070 mol HCl

        ⇒ x = 0.00070 mol NaHCO₃

4. Convert 0.0070 mol NaHCO₃ to grams:

• Mass in grams = molecular mass × number of moles
• molecular mass NaHCO₃ = 84.007 g/mol
• Mass of NaHCO₃ = 84.007 g/mol × 0.00070 mol = 0.0588 g ≈ 0.059 g

The answer has two significant digits because the molarity (0.035 M) is reported with two signficant digits.