What chemical reaction is happening in this lab \"GRAVIMETRIC DETERMINATION OF C

Question

What chemical reaction is happening in this lab "GRAVIMETRIC DETERMINATION OF CALCIUM AS OXALATE"? what are the chemical reactions??? What are the steps for finding the molarity of Ca 2+ in the unkown solution??? (what do I need to know? and find inorder to calculate that?)

PROCEDURE

1. Using suction, clean three medium-porosity, sintered-glass crucibles with dilute nitric acid followed by a thorough distilled water rinse,weigh them.

2. Use a few small portions of unknown to rinse a 25 mL transfer pipet, and discard the washings. Transfer exactly 25 mL of unknown to each of three 250-400 mL beakers, and dilute each with 75 mL of 0.1 M HCl. Add 5 drops of methyl red indicator solution to each beaker. red below pH 4.8 and yellow above pH 6.0.

3. SLOWLY add 25 mL of 0.32 M ammonium oxalate solution to each beaker. Add 15 g of solid urea to each sample, cover it with a watchglass, and boil gently on a hot plate set to "low" for 30 min. until the indicator turns yellow

4. Filter each hot solution through a weighed funnel using suction. Add 3 mL of ice-cold water to the beaker, transfer the remaining solid to the funnel. Repeat this procedure with small portions of ice-cold water until all of the precipitate has been transferred. Finally, use two 10 mL portions of ice-cold water to rinse each beaker, and pour the washings over the precipitate.

5. Dry the precipitate, first with aspirator suction for 1 min, then in oven. Bring each filter to constant weight.

6. Calculate the molarity of Ca 2+ in the unknown solution.

Explanation / Answer

ANSWER:

Calcium from a 25 mL unknown sample is precipitated as calcium oxalate upon addition of ammonium oxalate:

Ca+2 + (NH4)2C2O4 ---------- CaC2O4(s) + NH4+

TheCaC2O4(s) is then washed, dried and weighed to constant weight.

From the mass of CaC2O4(s), we can solve for the amount of calcium present in the 25 mL sample. Eventually, we can solve for its molarity.

For example, the mass of CaC2O4(s) obtained is 0.5 g (Note: this is just hypothetical data).

Solving for mole CaC2O4(s):

MoleCaC2O4 = 0.5 g CaC2O4  x (1 mole CaC2O4 / 128 g)

  MoleCaC2O4 = 0.0039

Solving for mole Ca+2:

Mole  Ca+2 = 0.0039 CaC2O4   x (1 mole Ca+2 / 1 mole CaC2O4)

  Mole  Ca+2 = 0.0039

Solving for the molarity of Ca+2:

M = 0.0039 mole / 0.025 L

M = 0.156 mol/L

Do the same for the other 2 trials, then solve for the average molarity of  Ca+2.

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