Answer:
0.0347 moles of hydronium ions
Explanation:
The equation of the neutralization reaction between hydroxide and hydronium ions is given below:
H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)
From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.
The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:
Number of moles = mass / molar mass
Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol
Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol
Mass of magnesium hydroxide = 450 g = 0.45 g
Mass of aluminium hydroxide = 500 mg = 0.5 g
Moles of magnesium hydroxide = (0.45/58) moles
Moles of aluminium hydroxide = (0.5/78) moles
Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:
Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)
Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)
Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles
Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles
Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions
Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.
