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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: KI equil
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[Post New]24 Jan 2008 07:38:12 IST
    Subject: KI equil
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punnima (563)

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a std soln of I2in water contains 0.33 g of  I2 in one litre flask.more than this dissolve in a KI soln bcoz of the foll. equil.
                                   I2+I-I3-
A   0.1M  KI soln. actually dissolves 12.5g of iodine per litre most of which is converted to I3- Assuming tat conc of I2 in all saturated soln. is same,calc K for d above reacn.
VARSHA KRISHNAN....
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IIT is always a word which rises E thru' d body. But 2 achieve it U hav 2 drain out d entire E out of the body....

    
[Post New]24 Jan 2008 08:30:52 IST
    Subject: Re:KI equil
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Greatdreams (3220)

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No of moles of I2 in 0.33 gms of I2 = .33/254 = 1.30 * 10 -3 mol

So in 12.5 gms of I2 contains 12.5/254 = .049 mol of I2

So [I3-]  = .049 - 1.30 * 10 -3 = .048 M

We already know that [I2] = 1.3 * 10 -3 M

and [I-] = .1 - .048 = .052 M

So the K for this reacn = [I3-]/[I2][I-] = .048 / [(1.3 * 10 -3) * .052] = 710.05

I hope this is the answer.
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[Post New]24 Jan 2008 23:37:53 IST
    Subject: KI equil
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punnima (563)

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thanx a lot..........d ans is rite.
but i didnn noe how
thanx again
VARSHA KRISHNAN....
------------------------------------
IIT is always a word which rises E thru' d body. But 2 achieve it U hav 2 drain out d entire E out of the body....

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