Question

If 5.85mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00 mL of 0.1000 M HCl was added to 1.00 g of an antacid, how many moles of acid can the antacid counteract per gram?

answer in moles/gram and show all work please!

Answer 1

no of moles of NaOH = molarity * volume in L

                           = 0.1*0.00585 = 0.000585moles

no of moles of HCl = 0.1*0.02 = 0.002moles

if 0.002 moles of HCl left over the 0.000585 moles of NaOH after addition of antacid

0.002-0.000585 = 0.001415 moles

0.001415 moles of antacid

1gm of antacid counteract with 0.001415 moles of HCl