Cyanide recovered from the refining of gold ore can be determined indirectly by
ID: 1048377 • Letter: C
Question
Cyanide recovered from the refining of gold ore
can be determined indirectly by EDTA titration. A known
excess of Ni2 is added to the cyanide to form tetracyanonickelate(
II):
4CN + Ni ---> Ni(CN)4
When the excess Ni2 is titrated with standard EDTA,
Ni(CN)4 does not react. In a cyanide analysis, 12.7 mL of
cyanide solution were treated with 25.0 mL of standard solution
containing excess Ni to form tetracyanonickelate. The excess
Ni required 10.1 mL of 0.013 0 M EDTA for complete
reaction. In a separate experiment, 39.3 mL of 0.013 0 M EDTA
were required to react with 30.0 mL of the standard Ni
solution. Calculate the molarity of CN in the 12.7-mL sample
of unknown.
Explanation / Answer
Ni2+ + EDTA4- -----> Ni-EDTA 2-
Number of moles of EDTA reacted ,n = Molarity x volume in L
= 0.0130 M x 39.3 mL x 10-3 L/mL
= 5.11x10-4 mol
From the above reaction ,
1 mole of Ni2+ reacts with 1 mole of EDTA
5.11x10-4 mol of Ni2+ reacts with 5.11x10-4 mol of EDTA
So concentration of Ni2+ is = number of moles / volume in L
= 5.11x10-4 mol / (30.0x10-3 L)
= 0.017 M
Therefore the number of moles of Ni2+ reacted with EDTA is = number of moles of EDTA
= Molarity x volume in L
= 0.013 M x 10.1 mL x 10-3 L/mL
= 1.313x10-4 moles = n
The number of moles of Ni2+ added to CN- is , n' = Molarity x volume in L
= 0.017 M x 25.0x10-3 L
= 4.25x10-4 mol
So the actual number of moles of Ni+ reacted with CN- is , N = n' - n = 2.94x10-4 mol
4CN- + Ni ---> Ni(CN)4
From the above balanced reaction ,
1 mole of Ni reacts with 4 moles of CN-
2.94x10-4 mol of Ni reacts with 4x2.94x10-4 mol= 1.17x10-3 mol of CN-
So Molarity of CN , M = number of moles / volume in L
= (1.17x10-3 mol) / (12.7 mLx10-3L/mL)
= 0.0925 M
Therefore the molarity of CN in the sample is 0.0925 M
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