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Calculate K(c ) for the reaction KII(2) ...

Calculate `K_(c )` for the reaction `KI_I_(2) hArr KI_(3)`. Given that initial weight of `KI` is `1.326 g` weight of `KI_(3)` is `0.105 g` and number of moles of free `I_(2)` is `0.0025` at equilibrium the volume of solution is `1-L`.

Text Solution

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`{:(,KI(aq),+,I_(2)(g), hArr, KI_(3)(aq)),(,,,,"Mw KI=166",),(,,,,"Mw KI_(3)=420",),("moles at",1.326/166,,0,,0),("t=0",,,,,),("moles at",[1.326/166-0.105/420],,0.0025,,0.105/420),("equilibrium",,,,,),(,=7.738xx10^(-3),,0.0025,,2.5xx10^(-4)):}`
`:. K_(c )=([KI_(3)])/([KI])=(2.5xx10^(-4))/(7.738xx10^(-3))`, ( :' Volume of solution = `1L`)
`=0.032`
The free iodine should be in solid state because of the dissolved iodine with `KL(aq)` which is in excess.
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