Question

C: An antacid tablet (Al (OH)₃ ) was dissolved in 250.00 mL of 0.200 M HBr. 30.00 mL of the acid solution (contains remaining HBr) was titrated to endpoint with 15.00 mL of 0.1500 M . How much aluminum hydroxide was in the antacid tablet?

D: Suppose 0.500 g of an antacid containing ( Mg (OH)₂ ) was titrated with 30.00 mL of 0.250 M hydrochloric acid. Find the % Mg(OH)₂ in this antacid.

Answer 1

C:

Al(OH)₃ + 3HBr AlBr₃ + 3H₂O

250 mL of 0.200 M of HBr = 50 mmol of HBr

Molarity of the remaining HBr after aluminium hydroxide neutralization = (15.00 x 0.1500)/30 = 0.075 M

Therefore the moles of HBr remained after neutralization = 250 x 0.075 = 18.75 mmol

Thus the moles of HBr that Al(OH)₃ consumed = (50 - 18.75) = 31.25 = 31.25 x 10^-3 mol

From the above balance equation it is clear that 3 moles of HBr consume 1 mole of Al(OH)₃

Therefore, 31.25 x 10^-3 mol of HBr will consume = 31.25 x 10^-3/3 = 10.42 x 10^-3 mol Al(OH)₃

Molar mass of Al(OH)₃ = 78 g/mol

Therefore, the mass of aluminum hydroxide in the antacid tablet = 78 x 10.42 x 10^-3 = 0.813 g

                                                                                                                                      = 813 mg

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D:

30.00 mL of 0.250 M hydrochloric acid = 30.00 x 0.250 = 7.50 mmol = 7.50 x 10^-3 mol

Mg(OH)₂ + 2HCl MgCl₂ + 2H₂O

From the balanced equation it is clear that 2 mol HCl reacts with 1 mol of Mg(OH)₂

Therefore, 7.50 x 10^-3 mol HCl will react with = (7.50 x 10^-3)/2 = 3.75 x 10^-3 mol of Mg(OH)₂

Molar mass of of Mg(OH)₂ = 58.3197 g/mol

Therefore, the mass of Mg(OH)₂ in the tablet = 58.3197 x 3.75 x 10^-3 = 0.219 g

Hence, the % Mg(OH)₂ in this antacid = (0.219 x 100)/0.500 = 43.80 %