The volume in ml of 0.71 M HCl that would be required to neutralize 11.33 g of 2.03 M NaOH is 30.3 ml
<u><em>calculation</em></u>
Step 1 : write equation for reaction
NaOH + HCl →NaCl +H₂O
Step 2:find the volume of NaOH
volume=mass/ density
= 11.33 g/ 1.0698 g/ml =10.59 ml
Step 3: find the moles of NaOH
moles = molarity x volume in liters
molarity= 2.03 M=2.03 mol/l
volume in liters =10.59/1000 =0.0106 L
moles = 2.03 M x 0.0106 L =0.0215 moles
step 4: use the mole ratio to determine the moles of HCl
from equation in step 1 , NaOH:HCl is 1:1 therefore the moles of HCl is also 0.0215 moles
Step 5: find volume of HCl
volume= moles/ molarity
molarity =0.71 M =0.71 mol/l
=0.0215 moles /0.71 mol/l=0.0303 L
into ml =0.0303 x 1000=30.3 ml
