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The equilibrium constant for the followi...

The equilibrium constant for the following reaction at `25^(@)` C.
`CN^(-) + HAC [Given that `K_(a)(Hac) = 1.8 xx 10^(-5) mol//dm^(3), K_(a)(HCN) = 0.45 xx 10^(-9)]`

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The correct Answer is:
4
`CN^(-) + Hac For the given reaction, the equilibrium constant is given as
`K=([HCN][AC^(-)])/([CN^(-)][Hac])`
Multiplying, and dividing by `[H_(3)O^(+)]`, we get
`K=([HCN]^(-)[AC^(-)][H_(3)O^(-)])/([CN^(-)][Hac][H_(3)O^(+)]) =([H_(3)O^(-)][Ac^(-)])/([HAc]) = ([HCN])/([H_(3)O^(-)][CN^(-)])`
`=(K_(a)(HAc))/(K_(a)(HCN)) rArr K = (1.8 xx 10^(-5))/(0.45 xx 10^(-9)) = 4 xx 10^(4)`
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