If 5.35mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00...
Homework Answers
mol NaOH=(0.00535L)(0.1M)=0.000535mol
mol HCl=(0.02L)(0.1)=0.002mol
mol HCl neutralized by antacid: 0.002mol -0.000535mol =0.001465mol
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