Asked by Kibito
My Question is:
Commercial vinegar usually has a concentration of 5%, per volume. Calculate the [H3O+], if the density of acetic acid is 1.051 g/mL.
What I did was:
mass = V * density
mass = 0.05 * 1.051 = 0.05 g
moles = grams/molar mass = 0.05/60.05 = 8.3 * 10^-4
M = moles/V = 8.3 * 10^-4/0.1 = 8.3 * 10^-3 M
I can do the rest on my own but is this much correct?
Thanks
Commercial vinegar usually has a concentration of 5%, per volume. Calculate the [H3O+], if the density of acetic acid is 1.051 g/mL.
What I did was:
mass = V * density
mass = 0.05 * 1.051 = 0.05 g
moles = grams/molar mass = 0.05/60.05 = 8.3 * 10^-4
M = moles/V = 8.3 * 10^-4/0.1 = 8.3 * 10^-3 M
I can do the rest on my own but is this much correct?
Thanks
Answers
Answered by
DrBob222
I don't think so. I don't know why you converted anything to mass although it can be done that way but not by your calculation. (For example, the volume is not 0.05). You want the molarity and that is mols/L. You already have it as 5% w/v which means 5 g/100 mL or 50 g/L.
How many mols is in 50 g? That gives you the molarity (and note we never used the density). Then calculate (H3O^+) using Ka for acetic acid.
You CAN use the density this way.
mass = volume x density
mass = 100 mL x 1.051 = 105.1 grams for mass % of 5/105.1 = 0.04757% w/w.
Then 1.051 x 1000 mL x 0.04757 x (1/60.05) = ? M but this is far more complicated and more work than the above.
How many mols is in 50 g? That gives you the molarity (and note we never used the density). Then calculate (H3O^+) using Ka for acetic acid.
You CAN use the density this way.
mass = volume x density
mass = 100 mL x 1.051 = 105.1 grams for mass % of 5/105.1 = 0.04757% w/w.
Then 1.051 x 1000 mL x 0.04757 x (1/60.05) = ? M but this is far more complicated and more work than the above.
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