Question

 

DATA SHEET

	Titration 1	Titration 2	Titration 3
Mass of KHP, grams	.10 g	.10 g	.10 g
Initial volume of NaOH, ml	0.0	6.2	12.1
Final volume of NaOH, ml	6.2	12.1	17.5

CALCULATIONS AND RESULTS

1. number of moles of KHP: n = mass KHP x (1 / 204 g/mol)

n ₁   = ______________        n₂   =_____________        n₃   =___________________

2. number of moles of NaOH = number of moles of KHP

n ₁   = __________________        n₂ =___________________       n₃   =______________________

3. Volume of NaOH used in each titration = final volume of NaOH - initial volume of NaOH

V ₁ =   _________________        V₂  =___________________        V₃   =_______________________

4. Molarity of NaOH solution = n/V

M ₁ =   _________________        M₂ =___________________       M₃   =______________________

2. Average molarity of NaOH:

M = (M₁ + M₂ + M₃) / 3 = ________________________________________

Answer 1

1.

Number of moles of KHP

n1 = (0.1 g) / (204 g/mol) = 0.0005 mol

n2 = (0.1 g) / (204 g/mol) = 0.0005 mol

n3 = (0.1 g) / (204 g/mol) = 0.0005 mol

2.

Since, Number of moles of NaOH = number of moles of KHP

So,

n1 _(NaOH)= n1 _(KHP) = 0.0005 mol

n2 _(NaOH)= n2 _(KHP) = 0.0005 mol

n3 _(NaOH)= n3 _(KHP) = 0.0005 mol

3.

Volume of NaOH used in each titration = final volume of NaOH - initial volume of NaOH

V1 = 6.2 mL – 0.0           = 6.2 mL = 0.0062 L

V2 = 12.1 mL – 6.2 mL = 5.9 mL = 0.0059 L

V3 = 17.5 mL – 12.1 mL = 5.4 mL = 0.0054 L

4.

Molarity of NaOH solution = n/V

M1 = (0.0005 mol) / (0.0062 L) = 0.081 M

M2 = (0.0005 mol) / (0.0059 L) = 0.085 M

M3 = (0.0005 mol) / (0.0054 L) = 0.093 M

5.

Average molarity of NaOH:

M = (M₁ + M₂ + M₃) / 3
    =[ (0.081 M) + (0.085 M) + (0.093 M) ] / 3

    = 0.259 M / 3

    = 0.086 M