Need some help with this, Ive done all the calculations but Im sure its wrong as

Question

Need some help with this, Ive done all the calculations but Im sure its wrong as vinegar is supposed to have around 4% acetic acid, and I got 49.8%.

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Initial NaOH reading

Final NaOH reading

Volume of NaOH used

Trial 1

9.6mL

1.4mL

8.2mL

Trial 2

9.7mL

1.5mL

8.2mL

Trial 3

9.8mL

1.4mL

8.4mL

Average volume of NaOH used:

9.7mL

1.4mL

8.3mL

Calculate the Normality of the vinegar using the previously given equation.

Na = (Nb)(Volumeb) / (Volumea)

N HC2H3O2 = (0.5 M NaOH) (8.3mL NaOH) / (5mL HC2H3O2)

= 8.3 M HC2H3O2

Calculate the mass of the acetic acid in grams using the previously given equation.

Massa = (Na)(GMWa)

Mass HC2H3O2 = (8.3 M HC2H3O2) (60.05 g/mol HC2H3O2)

= 498 g

Calculate the percentage of acetic acid using the previously given equation.

% Acid = (Massa (g/L) / 1000g/L) x 100

% Acid = (498 g/L / 1000 g/L) x 100 =

= 49.8 % acetic acid

Thanks!

<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /?>

Initial NaOH reading

Final NaOH reading

Volume of NaOH used

Trial 1

9.6mL

1.4mL

8.2mL

Trial 2

9.7mL

1.5mL

8.2mL

Trial 3

9.8mL

1.4mL

8.4mL

Average volume of NaOH used:

9.7mL

1.4mL

8.3mL

Explanation / Answer

If you look at your second equations: .5M(8.3mL)/5mL it would equal .83 not 8.3. Maybe this is where your error is so going through the rest of the equations your end result would be 4.98%

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