A cyanide solution with a volume of 22.00 mL was treated with 25.00 mL of Ni2 (c

Question

A cyanide solution with a volume of 22.00 mL was treated with 25.00 mL of Ni2 (containing excess
Ni^2+) to convert the cyanide into tetracyanonickelate(II) ion:
4 CN- (aq) + Ni^2+ (aq) -> Ni(CN)4 ^2-(aq)
The remaining unreacted Ni^2+ was reacted with 9.00 mL of 0.01500 M EDTA^4- (a reagent that forms 1:1
complexes with Ni2+ ion but does not react with Ni(CN)4^2- ).
EDTA4- (aq) + Ni2+(aq) -> Ni(EDTA)^2-
If 40.25 mL of the same EDTA titrant is needed to completely react with 35.10 mL of the original Ni^2+ solution, calculate the molarity of the CN- in the 22.00 mL sample.

Explanation / Answer

First, find the moles of EDTA in the 40.25 ml...
moles EDTA = 0.04025 * 0.015 = 0.00060375 moles of EDTA
And this moles react with 35.10 ml of Ni+2:
0.00060375 moles of EDTA / 35.10 ml Ni+2 =
1.72x10-5 moles EDTA / ml Ni+2

Now 25 ml of Ni+2 should require:
25*(1.72x10-5 moles EDTA / ml Ni+2) = 4.3x10-4 moles of EDTA

Now for the titrating moles
0.009 L EDTA * 0.015 mol/L = 1.35x10-4 moles EDTA

wich means that the moles of EDTA equivalent to the moles of Ni+2 that reacted with the CN was:
(4.3x10-4) - (1.35x10-4) moles of EDTA= 2.95x10-4 moles EDTA = moles of Ni2+

moles of CN = 2.95x10-4 moles * 4 moles CN / 1 mol Ni2+ = 1.18x10-3 moles of CN

Finally the molarity of CN:

1.18x10-3 moles / 0.022 L

M = 0.0536 M

Hope this helps

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