- #1
salman213
- 302
- 1
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?
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?