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?

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