I am having issues getting this calculations for my titrations: My solution was
ID: 1039863 • Letter: I
Question
I am having issues getting this calculations for my titrations:
My solution was calcium chloride from solid calcium chloride dihydrate and distilled water
we titrate 0.02 M calcium chloride solution
I used 0.2938 solid calcium chloride dihydrate
It was taken 10.0 mL of the solution.
and 70 mL of the titrate was used
my titrate was EDTA4-
and we use a buffer with ethanolamine,magnesium,sulfatem and EDTAH4
I need to find from each titration:
1) Moles of Ca2+ in 10.00 mL
2) Moles of EDTA4- in 10.00 mL
3) [EDTA4-]
3a) average [EDTA4-]
4) %RSD for [EDTA4-]
Even just a sample one will be awasome thank you.
Starting Ending Trial 1 10.0 mL 32.4mL Trial 2 0 mL 24.0mL Trial 3 0 mL 23.5mLExplanation / Answer
you have taken 0.2938 gm of solid calcium chloride dihydrate i.e. about 2 mmoles.
Since you have mentioned the concentration to be 0.02 M thus volume of your calcium chloride solution should be 100 ml. Of this 100 ml, 10 ml of solution has been subjected to titration.
Your titrant is EDTA4- which is a hexadentate ligand and combines with Ca2+ to form 1:1 complex thus mmoles of Ca2+ = mmoles of EDTA4-.
In the table you have given the data which i supposed must be titre value.
Trial 1;
In100 ml of solution mmoles of CaCl2.2H2O = 2 mmoles , then in 10 ml it would be 0.2 mmoles, and since mmoles of Ca2+ = mmoles of EDTA4- thus mmoles of EDTA4- = 0.2 mmoles.
Volume of EDTA4- used is 22.4 ml
Thus, Molarity * volume = moles
Hence, Molarity of EDTA4- = 0.2 / 22.4 = 0.0089 M
From Trial 2, volume = 24 ml
Hence, Molarity of EDTA4- = 0.0083 M
From Trial 3, volume = 23.5 ml
Hence, Molarity of EDTA4- = 0.0085 M
Average [EDTA4-] = 0.0086 M
Now moles of EDTA4- in 10 ml = M * 10 = 0.086 M
% RSD based on this calculation is 3.5%.
Hope this helps. Please share in comments.
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