Question

 Part A: Standardization of NaOH solution.

 Part B: Determination of the Molar Mass of a Solid Acid

 Unknown Number _______  Number of moles of ionizable H' ions per mole of solid acid Nur

 M_NaOH _______  (The average molarity of NaOH from Part A.)

Answer 1

Hii

Standardization of NaOH

Trial 1 –

Assumed molarity of HCl= 1M

Reaction between the NaOH and HCl

NaOH+HCl= NaCl+H20

Moles of NaOH reacted = moles of HCl reacted

Volume of NaOH used *molarity of NaOH= Volume of HCl*molarity of HCl

Molarity of NaOH= (Volume of HCl*molarity of HCl)/ Volume of NaOH used

Trial 1	Trial 2	Trial 3
Volume of HCl used = 32 ml	Volume of HCl used = 42.5 ml	Volume of HCl used = 50 ml
Volume of NaOH solution =34.3	Volume of NaOH solution =45.6	Volume of NaOH solution =50
Molarity of NaOH= 32*1/34.1=0.938	Molarity of NaOH= 42.5*1/45.6=0.932	Molarity of NaOH= 50*1/50 =1

Average molarity=( 0.938+0.932+1)/3= 0.956 M

I have assumed the molarity of HCl as 1 ,please ask your lab assistant and follow the same procedure incase the molarity used is different in your experiment.

Estimation of molecular mass-

You can assume unknown acid as monoprotic HA (having ability to donate one H+) because the ionization constant for the second H+ is very less and thus H+ contribution can be neglected

Reaction can be written as

HA+NaOH=NaA+H2O

Moles of HA= moles of NaOH

Moles of HA= molarity*volume in litre

1 ml=0.001 ml

Moles = mass of HA in gram/molecular mass in gram/mole

molecular mass in gram/mole=mass of HA in gram/moles of HA

Trial -1	Trial -2
Mass of sample =0.202 g	Mass of sample =0.229 g
Volume of NaoH consumed = 12.5-1=11.5 ml	Volume of NaoH consumed =23.5-12.5= 11 ml
Moles of HA=0.00115*1	Moles of HA=0.0011*1
Molecular mass of HA=0.202/0.00115=175.6 g/mole	Molecular mass of HA=0.229/0.0011=208.1 g/mole

Part b follows from the answer of part a where I assumed the molarity of HCl as 1,in case the molarity is different answer will be different.

You can always follow the same procedure in case the molarity is different or you can comment the molarity in comment box below answer, I will do calculations according to that and send

Thank you