0.688 g of antacid was treated with 50.00 mL of 0.100 M HCl. The excess acid required 4.21 mL of 1.200 M NaOH for back titration. What is the neutralizing power of this antacid expressed as mmol(millimoles) of HCl per gram of antacid?

Question

0.688 g of antacid was treated with 50.00 mL of 0.100 M HCl.  The excess acid required 4.21 mL of 1.200 M NaOH for back titration.  What is the neutralizing power of this antacid expressed as mmol(millimoles) of HCl per gram of antacid?

DrBob222 · Answer

mmols HCl added initially = mL x M = 5.00 mmols NaOH to neutralize = 4.21 x 1.2 = 5.05. Check you numbers. mmols NaOH > mmols HCl which means antacid tablet did not neutralize any of the HCl.