Question

Sol. to make 0.120 (M) HCl stock soln = 0.103 M Part A #1 | HCl (mL) NaOH (mL) initial 0.00 0.15 final 26.12 27.60 Endpoint was a clear, very faint pink solution. Part A #2 | HCl (mL) NaOH (mL) initial 0.48 0.40 final 25.55 27.16

Answer 1

This is an ACID-BASE neutralization reaction:

HCl (aq) + NaOH (aq) -------------------> H2O (l) + NaCl (aq)

a) Volume of HCl in titration#1 = Vfinal - Vinitial  = 25.55 mL - 0.48 mL = 25.07 mL =0.02507 L

b) Molarity = number of moles / volume of solution in L

Molarity of HCl = 0.103 M ,  number of moles of HCl=Molarity of HCl x volume of solution in L

= 0.103 M x 0.02507 L = 0.002582 moles =2.58 mmoles.

c) According to stoichiometry of the reaction, n moles of HCl reacts with n moles of NaOH, to form n moles each of water and NaCl.

Therefore, number of moles of NaOH=number of moles of HCl= 0.002582 moles =2.58 mmoles.

d) number of moles of NaOH=number of moles of HCl

but  number of moles = Molarity x volume of solution in L

therefore: MHCl x VHCl = MNaOH x VNaOH

MNaOH = ( MHCl x VHCl )/VNaOH

  VNaOH= Vfinal - Vinitial  =27.60 mL - 0.15 mL = 27.45 mL = 0.02745 L

MNaOH = ( MHCl x VHCl )/VNaOH = ( 0.103 M x 0.02507 L ) / 0.02745 L = 0.094 M