Question

1. If 35ml of 0.15 M HCl was used to neutralize 25 mL of a calcium hydroxide solution, what was the concentration of the base?

Answer 1

Reaction:

Ca (OH)₂ + 2HCl --> CaCl₂ + H₂O

V (HCl)=35ml=0.035 L

Mole n (HCl)=Molarity×Volume=

0.15M×0.035L=0.00525mol

n(HCl)=1/2n(Ca (OH)₂= 1/2×0.00525mol=0.002625mol

Volume(Ca(OH)₂)=25ml=0.025L

Molarity (Ca(OH)₂ )=

mole(Ca (OH)₂)/Volume (Ca (OH)₂)=

0.002625mol/0.025L=0.105 mol/L