Equilibrium Position - Effect of Dilution

[salman213]

Iodine is sparingly soluble in pure water. However, it does `dissolve' in solutions containing excess iodide ion because of the following reaction:

I-(aq) + I2(aq) I3-(aq) K = 710 L/mol
For each of the following cases calculate the equilbrium ratio of [I3-] to [I2]:

2.00×10-2 mol of I2 is added to 1.00 L of 2.00×10-1 M KI solution.

I did this question using the ICE table and got the RIGHT answer of

1.28×10^2

but now the SECOND part asked.. same thing but

The solution above is diluted to 5.50 L.


How do i approach this?

 

[salman213]

nevermind i got it :)

 

[rocomath]

did you re-use the value you found?

 

[salman213]

not really, i just made did the following

C1V1=C2V2
(2.00×10-1 M)(1.00 L) = C2(5.50L)

solved for c2 and then

2.00×10-2 mol of I2 is added to 5.50 L of C2 KI solution.

then again did ICE table and solved at equilibrium.. then did the same thing as part a to find ratio of I3 to I2 at equilibrium

 

[palsword]

Same question

I have the same question but with different numbers
can you show me how to solve it

8.00×10-2 mol of I2 is added to 1.00 L of 8.00×10-1 M KI solution.

The solution above is diluted to 13.00 L.

 

[rocomath]

M_{1} \times V_{1} = M_{2} \times V_{2}

solve for final Molarity

 

[salman213]

yea just solve for molarity first


(8.00x10^-1)(1.00L) = (c2)(13.00)

solve for c2 now ur new question is

8.00×10-2 mol of I2 is added to 13.00 L of c2 M KI solution.



make Ice table now so u have
...I-(aq)...I2(aq)...I3-(aq)
I...0.08/13...c2....0
C...-x....-x...x
E

x
--------------------- = k(watever k value ur given)
((0.08/13)-x)((c2) - x)

solve for ur x using quadratic formula or if ur lazy find one online and just type in a,b and c values to find x, then plug back into get
I3:I2

I3...x
-- = --------------
I2...c2-x