Question

Consider the reaction: HCl(aq) + NaOH(aq) → NaOH(aq) + H2O(l) How much 0.113 M NaOH solution will completely neutralize 1.25 L of 0.228 M HCl solution?

Answer 1

1.25L Hcl(.228 mol HCl/1 L)=0.285 mol HCl
0.285 mol HCl(1 mol NaOH/1 mol HCl)=0.285 mol NaOH
0.285 mol NaOH(1 L/.113 mol NaOH)=2.52212 L of .113M NaOH solution

Answer 2

The  much of 0.113 M NaOH  solution that  is required  to  neutralize  is  2.522 L of NaOH

    calculation

HCl (aq) + NaOH (aq) → NaOH (aq)  + H2O  (l)

Step 1:  calculate the   moles of  HCl

  moles= molarity  x volume in liters

volume in liters =  1.25 L

molarity =  0.228 M  =  0.228 mol/ L

moles is therefore = 0.228 mol/L  x 1.25 L = 0.285  mols

Step  2:  use the mole ratio to  calculate  the   moles of NaOH

The  mole ratio   of  HCl : NaOH  is 1:1  therefore  the moles  of NaOH= 0.285 moles

Step  3:  calculate the volume  of  NaOH

volume=moles/molarity

molarity  0.113 M  = 0.113 mol/l

= 0.285 moles /0.113 Mol/  L = 2.522 L  of NaOH