It takes 19.72 mL of a 0.0225 M solution of EDTA to titrate the zinc present in 19.97 mL of a stock solution containing 0.316 g zinc chloride in 100.0 mL. What is the mass of zinc titrated?
The answer was .0290 grams.
...HOW?! lol
at the bottom of the answer sheet, it said:
Calculate the #moles of EDTA used in the titration:
vol (L) x [EDTA] (mol/L)
Since one mole of EDTA complexes one mole of Zn, that #moles x molar mass Zn = mass of zinc in sample
only 3 sig fig allowed because concentration of EDTA known to 3 sig fig
HELP!!
best answer
The answer was .0290 grams.
...HOW?! lol
at the bottom of the answer sheet, it said:
Calculate the #moles of EDTA used in the titration:
vol (L) x [EDTA] (mol/L)
Since one mole of EDTA complexes one mole of Zn, that #moles x molar mass Zn = mass of zinc in sample
only 3 sig fig allowed because concentration of EDTA known to 3 sig fig
HELP!!
best answer
-
It takes 19.72 mL of a 0.0225 M solution of EDTA to titrate the zinc present in 19.97 mL of a stock solution containing 0.316 g zinc chloride in 100.0 mL. What is the mass of zinc titrated?
EDTA = Ethylenediaminetetraacetic acid. Write the formula as C10H16N2O8 and add , to get a molecualer weight of 292.24 g/mol
When you titrate, the number of existing moles (equivalents) is equal to the number of the equivalents of EDTA I need in order to combine with all the equivalents of Zn compound.
# of equivalents I use = volume used x molarity (or normality)
so V(EDTA) x M (EDTA) = V (Zn) x M (Zn)
but here, as you have the volume used, you say:
in 100.0 ml of solution -------- 0.316 g zinc chloride
in the 19.97 mL used, -------- x, which you calculate to be 0,0631g of ZnCl2
how much Zn in a mole?
Zn 65.4
2 Cl 2x 35.5 so ZnCl2 has a molecular weight of 136,4 of which 65.4 is Zn
so if in a mass of 136,4 g of ZnCl2 there are 65.4 g of Zn
in 0,0631g of ZnCl2 we have x = I hope u get it!
EDTA = Ethylenediaminetetraacetic acid. Write the formula as C10H16N2O8 and add , to get a molecualer weight of 292.24 g/mol
When you titrate, the number of existing moles (equivalents) is equal to the number of the equivalents of EDTA I need in order to combine with all the equivalents of Zn compound.
# of equivalents I use = volume used x molarity (or normality)
so V(EDTA) x M (EDTA) = V (Zn) x M (Zn)
but here, as you have the volume used, you say:
in 100.0 ml of solution -------- 0.316 g zinc chloride
in the 19.97 mL used, -------- x, which you calculate to be 0,0631g of ZnCl2
how much Zn in a mole?
Zn 65.4
2 Cl 2x 35.5 so ZnCl2 has a molecular weight of 136,4 of which 65.4 is Zn
so if in a mass of 136,4 g of ZnCl2 there are 65.4 g of Zn
in 0,0631g of ZnCl2 we have x = I hope u get it!
-
I will start you out. What many people don't realize is that a 1 M solution also means 1 mmole per mL, and you know how many mL you used. So now you can calculate how many mmoles of EDTA you used.