Question

Titration of Vinegar Lab Report 1 /25 pts) Name: Part 1: Data and Calculations mass KHC.H.O. - 4.0744 g Molecular Weight potassium hydrogen phthalate KHC.H. - [1 pt) #moles KHP = [0.5 pts) Volume of solution = 250 ml_ L10.5 pts] Concentration of KHP = Mop = _M (1 pt) Volume of KHP Solution to be titrated mL = [1 pt) #mols of KHP to be titrated [Mc- XV] - __ [1 pt) Rough Trial Trial 1 Trial 2 0.0 9.9 0.0 Initial Volume of NaOH (+0.1 mi) Final Volume of NaOH (+0.1 ml) Volume of NaoH added (+0.1 mi) 99 20.9 10.4 11.0 10.4 [1 pt] Average Volume of NaOH added in 3 trials (0.1 mL) "Initial volume is the initial reading of the burette and final volume is the reading after adding NOOH solution [4 pts) Use the average volume of NaOH added (VNCH) to determine the molar concentration of NaOH used in the titration. Use the equation which applies to the situation that exists at the end point: M V . M .V . More specifically since KHP is an acid and sodium hydroxide is a base, M V MNOHVHow should be used and solve for M

What information then do you need? This is Chem 2 and not Org Chem.

Answer 1

Part 1

Molecular weight potassium hydrogen phthalate KHC8H4O4 = 204.22 g/mol

Moles of potassium hydrogen phthalate (KHP) = Mass of KHP/ Molecular mass = 4.0744/204.22 = 0.02 moles

Volume of solution= 250 mL= 0.25 L

Concentration of KHP= MKHP= Number of moles/volume in litre= 0.02/0.25 = 0.08 mol/L or 80 millimoles/L.

Volume of KHP solution to be titrated = VKHP= 250 mL = 0.25 L

Moles of KHP solution to be titrated = MKHP X VKHP = 0.08 X 0.25 = 0.02 moles

Average volume of NaOH added in three trial ( $\pm$ 0.1 mL) = 10.43 mL

The molar concentration of NaOH used in the titration using average volume by the equation

MacidVacid = Mbase Vbase

now putting the values we have a x .0.01043 L= 0.08 x 0.25 L

Now a = 0.08 x 0.25 L/ 0.01043 L = 1.917 mol/L

Part 2

The volume of NaOH dispensed in trial 1, 2 and 3 is 18.65 mL, 18.55 mL and 18.6 mL respectivelyv and the mean volume of NaOH dispensed is 18.6 mL. and the molar concentration of the NaOH standardized i n Part 1 of the procedure is 1.07 mol/L.

The molar concentration of acetic acid in undiluted vinegar is 50/60 =0.833 mol/L. Since undiluted vinegar contains 50 g of acetic acid in 1 L and molecular weight of acetic acid is 60.

Therefore 1 L of undiluted vinegar sample = 0.833 mol/L of acetic acid

using 1:10 dilution of vinegar we have MvinegarVvinegar = MNaOH VNaOH = Mvinegar x 0.1 L = 1.91 x 0.0186 L

Mvinegar = 1.91 x 0.0186 L/ 0.1 L = 0.35526 moles/L