Question

Given the NaOH density =1.0698g mL^-1, How many mL of .71M HCl would be required to neutralize 11.33g of 2.03 M NaOH?

Answer 1

 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

calculation

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