Calculate the molarity M of NaOH solution which required 22

Calculate the molarity (M) of NaOH solution which required 22 15 mL of react completely with 0 552 g KHP This problem is similar to what done in the Lab and calculation steps shown (Part A) Show all steps by using convert factors. Must finish with c.v. and f.v. Calculate the volume (mL) of 0 450 M of KOH required to neutralize 25.55 mL of 0.155 M HCI solution. Show Equation and rearranged Equation before solving the problem Show all steps with conversion factor and formula, equations etc.to get full credit.

Solution

HKC8H4O4

no of moles of KHP = W/G.M.Wt    = 0.552/204 = 0.0027 moles

HKC8H4O4 + NaOH --------> NaKC8H4O4 + H2O

1 mole            1 mole

1 mole of KHP react with 1 mole of NaOH

0.0027 moles of KHP react with = 1*0.0027/1 = 0.0027 moles of NaOH

molarity of NaOH = no of moles /volume in L

                                = 0.0027/0.02215   = 0.122 M

2. KOH + HCl ---------> KCl + H2O

    1 mole    1 mole

    KOH                                     HCl

M1 = 0.45M                             M2 = 0.155M

V1 =                                         V2 = 25.55ml

n1 = 1                                       n2 = 1

           M1V1/n1    =    M2V2/n2

              V1            = M2V2n1/n2M1

                               = 0.155*25.55*1/1*0.45 = 8.8ml >>> answer