Given moles EDTA = Molarity of EDTA x Volume EDTA (in L) and MOles of Ni2+ moles

Question

Given moles EDTA = Molarity of EDTA x Volume EDTA (in L) and MOles of Ni2+ moles EDTA

moles Ni2+/g sample = mole Ni2+/g

(Averagel moles NH3/g)/ (Average moles Ni2+/g) = x

If the sample of the complex ion compound that you prepared had not thoroughly dried, but contained traces of water or ethanol would this make your determination of the moles NH3/g and Moles Ni2+/g too high, too low, or would it not affect your results at all? Would this make your determination of the value of x too high, too low or would it not affect your results at all? Explain

Explanation / Answer

This will help you out.

Equation of reaction:
Ni2+ + EDTA4- ? NiEDTA2-
Mass of sample = 0.502g
V of final soln = 100ml
V of aliquot = 10mL
Conc of EDTA = 0.0101M
Results:
Titre 1 = 19.45 - 0.00 = 19.45ml
Titre 2 = 38.65 - 19.45 = 19.20ml
Titre 3 = 19.15 - 0.00 = 19.15ml
Titre 4 = 38.25 - 19.15 = 19.10ml

Using only titre values 1 -- 3,
Avg titre value = (19.20 + 19.15 + 19.10)/3 = 57.45/3 = 19.15ml

Ni2+/EDTA = 1:1
CnVn = CeVe
Cn = CeVe/Vn
= 0.0101 x 19.15/10 = 0.01934M
Conc of Ni2+ = 0.01934M
In 100ml (= 0.1L) soln, 0.01934M x 0.1L = 0.001934moles of Ni2+

Mw of Ni = 58.7g/mol
Mass of Ni2+ in salt = 0.001934 x 58.7 = 0.1135345g
0.502g of sample dissolved in 100ml solution gives % of Ni2+ as;
0.1135345/0.502 x 100 = 22.6%
%Ni2+ = 22.6%

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