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