To calculate the mass percent of acetic acid in the cleaning vinegar, we need to first calculate the molarity of the cleaning vinegar. We can do this by using the following equation:
Drops vinegar x M = Drops NaOH x M
10 drops vinegar x M = 25 drops NaOH x 0.0683M
M vinegar = 0.2732M
Now, we can calculate the mass percent of acetic acid in the cleaning vinegar using the following equation:
Mass percent acetic acid = (Molarity acetic acid x Molecular weight acetic acid) / (Density vinegar x 1000)
Mass percent acetic acid = (0.2732M x 60.05 g/mol) / (1.005 g/ml x 1000)
Mass percent acetic acid = 16.3%
Trying to find mass percent of acetic acid in 25 drops of .683M NaOH solution are required to neutralize 10 drops of cleaning vinegar. Assume that the density of the cleaning vinegar is the same as the density of regular vinegar which is 1.05 g ml-1
You need to clean up the post some. As it now reads, "Trying to find the mass percent of acetic acid in 25 drops of .683M NaOH solution are required to neutralize 10 drops of cleaning vinegar." I doubt there is ANY acetic acid in .683M NaOH. I assume it should read, "Trying to find the mass percent of acetic acid. 25 drops of 0.683M NaOH are required to neutralize 10 drops of cleaning vinegar.
drops vinegar x M = drops NaOH x M
Solve for M vinegar.
Now, do you know the size of a drop? or the mass of the drop? Or can you find percent from the molarity and the density?
A nationally known company markets a product called "cleaning vinegar" that is not designed for human consumption. What is the mass percent of acetic acid in this product, if 25 drops of .0683M NaOH solution are required to neutralize 10 drops of the cleaning vinegar? Assume that the denisty of the cleaning vinegar is the same as the density of regular vinegar, which is 1.005 g ml^-1.
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