Lab Experiment #625
Estimating the Acid Content of Fruit Juices
Another titration, you say? Oh boy, oh boy.
In this one we estimated the acid content of a given fruit juice by titrating it with a standardized sodium hydroxide solution. We assume that all of the acids are citric acid, or H3C6H5O7.
The equation thus looks like this:
H3C6H5O7(aq) + 3NaOH(aq) ----------> Na3C6H5O7(aq) + 3H2O
So the problem given is to...
...determine the number of grams of acid in one mL of an orange juice sample, assuming that the acid content consists entirely of H3C6H5O7. A titration of 10.0mL of the orange juice required 39.62mL of 0.106M NaOH.
1. Caluclate the number of moles of NaOH required to neutralize the acid. We remember from previous titrations that # of moles is equal to concentration multiplied by volume.
So let's convert the volume first so we can plug it in properly, because REMEMBER! All volume must be in LITERS....
39.62mL x 1L/1000mL = 3.962 x 10^-2mL so we can plug that into the equation:
number of moles = (3.962 x 10^-2 L)(0.106mol/L) = 0.4199x10^-2, adjust notation: 4.20 x 10^-3 mol NaOH
2. Calculate the number of moles of H3C6H5O7 in the titrated sample.
By looking at our given chemical equation we can see that 1 mole of acid reacts with 3 moles of NaOH, so the ratio is 1:3. This will become part of our equation.
Number of moles of acid = #moles of NaOH X 1 mol acid/3 mol NaOH (the molar ratio),
# moles H3C6H5O7 =
(4.20 x 10^-3 mol NaOH)(1mol H3C6H5O7/3 mol NaOH) = 1.4 x 10^-3 moles H3C6H5O7
3. Calculate the mass of H3C6H5O7 in the juice sample. The molar mass of H3C6H5O7 is 192.12g/mol.
so basically we need to convert the moles we found into grams:
1.4x10^-3 mol H3C6H5O7 = 192.12 grams/1mole = 0.269 grams H3C6H5O7
4. Calculate the mass of H3C6H5O7 in one mL of the juice.
So the to get mass of H3C6H5O7 in one mL, we are doing grams per mL calculations, basically just the mass of the H3C6H5O7 divided by the volume of the juice titrated in mL,
0.269g H3C6H5O7/10.0mL juice = 2.69 x 10^-2g H3C6H5O7/mL juice