hi, i really need help with this problem. i found that: the standardized molarit

Question

hi, i really need help with this problem.

i found that:

the standardized molarity of the EDTA is [0.0083] M

the molarity of the unknown Ca is [0.3204] M

i really need help with calculating the ppm of the Ca(unknown) and the ppm of the CaC03.

below is the question i am trying to solve.

i will not rate any "bad" answers or copied and pasted posts from wiki, if you can show your calculation that would also be a great help.

thank you for your help in advance!

You titrate a Ca containing unknown solution with the standardized EDTA. Given the following data:

f) ppm of CaCO3 in the unknown

___ ppm CaCO3

Volume of unknown titrated 5.00 mL Volume of EDTA used 19.19 mL Calculate the following: d) Molarity of Ca in the unknown 0.3204 M e) ppm Ca in the unknown ___ ppm Ca

f) ppm of CaCO3 in the unknown

___ ppm CaCO3

Explanation / Answer

ppm is defined as 1 gm of solute per liter of solution.

you haveconcentration of unknown Ca = 0.3204 M. This means 0.3204 mole Ca in 1 L of solution.

Thus you need to convert the mole number into mg

atomic weight of Ca is 40.1 g/mol

mass of Ca in 1 L = 0.3204 * 40.1 = 12.85 g

multiply this by 1000 to conert mass into mg

mass of Ca in 1 L = 12.85 * 1000 = 1285 mg

thus concentration in mg /L = 12850 mg/L

as 1mg/L = 1ppm therefore

concentration of Ca in ppm = 12850 ppm

now as molarity of CaCO3 is same as that of unknown Ca thereofre

mole of CaCO3 in 1 L = 0.3204 mol

molar mass of CaCO3 is 100.1 g/mol

mass of CaCO3 in 1L = 0.3204 * 100.1 = 32.07 g

convert mass into mg

mass of CaCO3 in 1L = 32.07 * 1000 = 32070 mg

as 1mg/L = 1ppm therefore

concentration of CaCO3 in ppm = 32070 ppm

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