Answer:
One tablet can neutralize 150 mL of stomach acid at a pH of 1.0
Explanation:
Step 1: Data given
Mass of calcium carbonate = 750 mg = 0.75 grams
Molar mass of CaCO3 = 100 g/mol
pH = 1.0
Step 2: The balanced equation
2HCl + CaCO3 → CaCl2 + H2CO3
Step 3: Calculate molarity of HCl
pH = -log[H+] = 1
[H+] = 0.1 M = 0.10 mol/L
Step 4: Calculate moles of CaCO3
Moles CaCO3 = 0.75 grams / 100g/mol
Moles CaCO3 = 0.0075 mol
Step 5: Calculate moles of HCl
For 2 moles HCl we need 1 mol CaCO3 to produce CaCl2 and 1 mole of H2CO3
For 0.0075 moles of CaCO3 we have 2*0.0075 = 0.015 moles HCl
Step 6: Calculate volume of HCl
Volume = moles /molarity
Volume = 0.015 moles / 0.1 M
Volume = 0.15 L = 150 mL
One tablet can neutralize 150 mL of stomach acid at a pH of 1.0.
