Cyanide ions can be determined by indirect complexometry. An excess of Ni 2+ is

Question

Cyanide ions can be determined by indirect complexometry. An excess of Ni2+ is added to

complex all CN- ions and the excess is then titrated par EDTA:

4 CN + Ni2+ -----> Ni(CN)4 , Kf= 10^30

Ni2+ + Y4- -----> NiY2-   Kf = 10^18

As the excess of nickel is titrated by EDTA (Y4-), the complex with cyanide does not react.

If 12.75 mL of an unknown CN- solution is treated with 25.00 mL of a 0.0140 M Ni2+ solution, the

excess of Ni2+ requires 10.21 mL of 0.012 M EDTA afterward. Determine the molarity of the

cyanide solution.

Explanation / Answer

excess moles of Ni2+ = moles of EDTA used = 0.012*0.01021 = .12252*10^-3 moles

moles oc Ni2+ consumed while reacting with CN- = 0.025*0.0140 -0 .12252*10^-3 = 0.227*10^-3

1 mole of Ni2+ reacts with 4 moles of CN-

hence moles of CN- = 4* 0.227*10^-3 = 0.909*10^-3 moles

hence concentrtion of CN - solution = 0.909*10^-3 / 0.01275 = 0.7129 M

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