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

Question

Cyanide ions can be determined by indirect complexometry. An excess of Ni2+ is added to complex all CN- ions and the excess is the titrated par EDTA:

4CN- + Ni2+ --> Ni(CN)4 2- kf=10^30

Ni2+ + Y4- ---> NiY2- kf = 10^18

As the excess of nickle is titrated by EDTA(y4-), the comples with cyanide does not react. if 12.75 mL of anunknown CN- solution is treated with 25 mL of a 0.0140 M Ni2+ solution, the excess of Ni2+ requires 10.21 mL of 0.012M EDTA afterward. Deermine the molarity of the cyanide solution.

Explanation / Answer

4 CN- + Ni2+ => Ni(CN)42-

Ni2+ + Y4- => NiY2-


Initial moles of Ni2+ = volume x concentration of Ni2+

= 25/1000 x 0.0140 = 0.00035 mol


Moles of excess Ni2+ = moles of EDTA = volume x concentration of EDTA

= 10.21/1000 x 0.012 = 0.00012252 mol


Moles of Ni2+ reacted with CN- = initial - excess moles of Ni2+

= 0.00035 - 0.00012252 = 0.00022748 mol


Moles of CN- = 4 x moles of Ni2+ reacted with CN-

= 4 x 0.00022748 = 0.00090992 mol


Molarity = moles/volume of CN-

= 0.00090992/0.01275

= 0.07137 M = 0.0714 M

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