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Calculate K(f) for the reaction: M^(3+...

Calculate `K_(f)` for the reaction:
`M^(3+) + SCN^(Theta) hArr MSCN^(2+)`,
The `[M^(3+)]` in the solution is `2.0 xx 10^(-3)M, [SCN^(Theta)] = 1.5 xx 10^(-3)M` and Free `[SCN^(Theta)] = 1.0 xx 10^(-5)M`.

Text Solution

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The correct Answer is:
A, B
`[SCN^(Theta)]` in complex formation `= (1.5 xx 10^(-3))-(1.0 xx 10^(-5))`
`=1.49 xx 10^(-3)M`
Hence, same is the concentration of `MSCN^(2+)` and bound `M^(3+)` concentration.
Thus `[M^(3+)]_(free) = (2.0 xx 10^(-3)) -(1.49 xx 10^(-3))`
` = 0.51 xx 106(-3)M`.
`K_(f) = ([MSCN^(2+)])/([M^(3+)][SCN^(Theta)]) =(1.49xx10^(-3))/((0.51xx10^(-3))(1.0xx10^(-5)))=2.9xx10^(5)`.
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