I am getting confused on how to do this problem .. In #1 is it .10g *1/ 204g/mol

Question

I am getting confused on how to do this problem .. In #1 is it .10g *1/ 204g/mol = 4.9 x 10^-4, so n1,n2, and n3 are all the same?

DATA SHEET

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 = ________________________________________

Explanation / Answer

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

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