Question

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.
For each of the following cases calculate the equilibrium ratio of [I3-] to [I2].

a) 9.00×10-2 mol of I2 is added to 1.00 L of 9.00×10-1 M KI solution
b) the solution above is diluted to 5.50 L

Answer 1

mols I2 = 0.0900; mols KI = M x L = 0.900 x 1.00 = 0.900
.......................I-(aq) + I2(aq)= I3-(aq)...................... K = 710
I......................0..........0.09........0
add................0.900...........................
C.....................-x............-x..........+x
E....................0.900-x....09-x........+x

K = (I3^-)/(I^-)(I2)
710 = (x)/(0.900-x)(0.0900-x)
Solve for x, evaluate (I3^-) and (I2), the the ratio.

For b, dilute the E line of problem a to 5.50 L, calculate the concentration of all the three species and use that as the I line and go through the same calculation shown for part a. Post your work if you get stuck.

Answer 2

so for a) I found the x=0.089 and plugged the x in the E line to find the equilibrium concentration. [I3]= 0.098, [I2]= 0.09-0.089= 2 x10^-4. To find the equilibrium ratio do I do [I3] / [I2]? If I do that I got 449 but the answer is 575. Im not sure what I did wrong. For part b how do I dilute the E line to 5.50? Do I times the E line by 5.50 or divide?

Answer 3

For a. I don't get your numbers. I think I3 is transposed and should be 0.089. I don't think 0.09 - 0.089 = 2E-4. I have a ratio of 890 if I go through the calculations but I rounded too much. The basic problem for both of our calculations is that both you and I rounded. You should carry out to enough places' i.e., at least to 3 significant figures and for small numbers like this and the quadratic formula I would do 4 places and round to three for the final number. There will be some purists that will argue that's not sound mathematically, and I know that, however, often that's the way the "book answer" is achieved.
For part b, first you need to make sure the E line is correct. Next you must note that those are moles we put in for I, C, E; however, those were for 1 L; therefore, those are molarities, After you get the E line squared away. You take the concentrations in molarity and divide by 5.50 L to get the new molarity. Thensubstitute those numbers into the equation for the I line (initial) and work the problem all over again. Good luck.