cyanide can be determined indirectly by EDTA titration. A known excess of Ni2+ i

Question

cyanide can be determined indirectly by EDTA titration. A known excess of Ni2+ is added to the cyanide to form tetracyanonickelate: 4CN- + Ni2+ = Ni(CN)4 ^2-

When exess Ni2+ is titrated with standard EDTA, Ni(CN)4 ^ 2- does not react. In a cyanide analysis 12.7 mL of cyanide solution was treated with 25.0 mL of standard solution containing excess Ni2+ to form tetra cyanonickelate. The excess Ni2+ required 10.1 mL of 0.013 M EDTA for complete reaction. In a separate experiment, 39.2 mL of 0.013 M EDTA was required to react with 30.0 mL of the standard Ni2+ solution. Calc. molarity of CN- in the 12.7 mL sample of unknown.

So far I have drawn a picture. And then I'm stuck

Explanation / Answer

Ni2+ + EDTA ---> (NiEDTA)2-

hence moles of Ni2+ = moles of EDTA

30 x M = 39.2 x 0.013 , M = 0.017 Molar Ni2+ ,

Ni2+ reacted with EDTA in first case = 10.1 x 0.013 = 0.1313 milli moles ( since vol is in ml)

total Ni2+ milli moles in first case = 25 x 0.017 = 0.425

Ni2+ reacted with CN- = 0.425-0.1313 = 0.2937

CN- milli moles = 0.2937 x4 = 1.1748 ( as per stochiometry in equation )

CN- molarity = 1.1748/12.7 = 0.0925 M

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