0.5358 grams of liquid antacid is mixed with 50 mL of DI water. An appropriate indicator and 180.00 mL of the HCl solution are added. If it requires 120.31 mL of the KOH solution to back titrate the excess HCl to the endpoint, how many moles of acid were neutralized by the antacid sample?
Assume that the moalrity of KOH and HCl is 0.100 M
The reaction of KOH and HCl ia as follows:
KOH+ HCl = KCl + H2O
First calculate the moles of KOH = Molarity * volume in L
= 0.100 *120.31/1000
= 0.012 Moles of KOH reacted with the same mole of HCl
Total mole of HCl = 0.100 *180/1000
= 0.0180 moles
Moles of HCl which are reacted with antacid = total moles – moles which are reacted KOH
= 0.0180 -0.012
= 0.006 Mole HCl
0.006 moles of acid were neutralized by the antacid sample. If the moalrity of KOH and HCl is 0.100 M
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