Question
How many grams of Al(OH)3 are required to neutralize 300mL of stomach acid HCl with a pH of 1.5?
Answers
DrBob222
pH = 1.5 = -log*(H^+)
(H^+) = 0.032 M
..............Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
millimoles HCl = mL x M = 300 mL x 0.032 = 9.6
The equation tells you that 1 mol Al(OH)3 requires 3 moles NaOH; therefore, 9.6 millimoles of HCl will require 9.6/3 = 3.2 millimoles AlOH)3 or 0.0032 mols Al(OH)3. Convert to grams by
grams Al(OH)3 = mols x molar mass = ?
Post your work if you get stuck.
(H^+) = 0.032 M
..............Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
millimoles HCl = mL x M = 300 mL x 0.032 = 9.6
The equation tells you that 1 mol Al(OH)3 requires 3 moles NaOH; therefore, 9.6 millimoles of HCl will require 9.6/3 = 3.2 millimoles AlOH)3 or 0.0032 mols Al(OH)3. Convert to grams by
grams Al(OH)3 = mols x molar mass = ?
Post your work if you get stuck.