Titration problem

Question

what vol of 0.100M NaOH solu nece to neutralize the excess HCl if 50.00ml of 0.121 M HCl are added to a 0.200g if Calcium carbonate

Answer 1

This is a back titration problem.  First, find out how much HCl is used to neutralize the calcium catbonate.   Then determine the excess HCl, and finally find the amount of NaOH to neutralize that.

1)  2HCl + CaCO3 ==> CaCl2 + H2O + CO2

moles CaCO3 present = 0.200 g x 1 mole/100 g = 0.002 moles

moles HCl used to neutralize this amount = 0.002 moles CaCO3 x 2 moles HCl/mole CaCO3 = 0.004 moles HCl used

moles of HCl actually present = 0.05 L x 0.121 mole/L = 0.00605 moles 

Excess moles of HCl present = 0.0065 moles - 0.004 moles = 0.00205 moles HCl in excess

2)  NaOH + HCl ==> NaCl + H2O

moles HCl present = 0.00205 moles

moles NaOH needed to neutralize this = 0.00205 moles (1:1 mole ratio of HCl : NaOH)

Volume of 0.100 M NaOH needed:  (0.100 mole/L)(x L) = 0.00205 moles

x = 0.0205 Liters = 20.5 mls of NaOH needed to neutralize the excess HCl