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
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.
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?
Why did the iodine go to school?
Because it wanted to make sure it passed the "I3" exam!
Let's find the equilibrium ratio of [I3-] to [I2] in each case:
a) In this case, we have 9.00×10-2 mol of I2 added to 1.00 L of 9.00×10-1 M KI solution. We can use the equation K = [I3-] / [I2] to solve for the equilibrium ratio.
First, let's find the concentration of I2:
[ I2 ] = (9.00×10-2 mol) / (1.00 L) = 9.00×10-2 M
Next, let's find the concentration of I3- using the equation K = [I3-] / [I2]:
710 = [I3-] / (9.00×10-2 M)
Simplifying the equation, we can find [I3-]:
[I3-] = 710 * (9.00×10-2 M) = 6.39 M
Therefore, the equilibrium ratio of [I3-] to [I2] is:
[I3-] / [I2] = 6.39 M / 9.00×10-2 M = 71.
b) In this case, the solution from part (a) is diluted to 5.50 L. The number of moles of I2 and KI remains the same, but the volume changes.
To find the concentration of I3- in the diluted solution, we need to adjust the volume:
[I3-] = [I3-] from part (a) / (1.00 L) * (5.50 L) = 6.39 M * 5.50 = 35.145 M
Therefore, the equilibrium ratio of [I3-] to [I2] is:
[I3-] / [I2] = 35.145 M / 9.00×10-2 M ≈ 390.5.
I hope these answers didn't go over your "I2" many details!
To calculate the equilibrium ratio of [I3-] to [I2], we need to use the equilibrium constant expression and the given information in each case.
a) 9.00×10^-2 mol of I2 is added to 1.00 L of 9.00×10^-1 M KI solution:
Step 1: Write the balanced equation for the reaction:
I-(aq) + I2(aq) = I3-(aq)
Step 2: Determine the initial concentrations of I- and I2:
[I-] = 9.00×10^-1 M
[I2] = 9.00×10^-2 mol / 1.00 L = 9.00×10^-2 M
Step 3: Use the given equilibrium constant (K = 710):
K = [I3-] / ([I-] * [I2])
Step 4: Substitute the initial concentrations into the equilibrium constant expression:
710 = [I3-] / (9.00×10^-1 M * 9.00×10^-2 M)
Step 5: Rearrange the equation to solve for [I3-]:
[I3-] = 710 * (9.00×10^-1 M * 9.00×10^-2 M)
[I3-] = 5.67×10^-2 M
Now we have the equilibrium concentration of I3-.
b) The solution above is diluted to 5.50 L:
Step 1: Since the solution is diluted, the moles of I2 and I- remain the same. However, the volume changes, so we need to recalculate the molar concentration of I2 and I-.
[I2] = 9.00×10^-2 mol / 5.50 L = 1.64×10^-2 M
[I-] remains the same at 9.00×10^-1 M.
Step 2: Use the same equilibrium constant expression and plug in the updated concentrations:
K = [I3-] / ([I-] * [I2])
710 = [I3-] / (9.00×10^-1 M * 1.64×10^-2 M)
Step 3: Rearrange the equation to solve for [I3-]:
[I3-] = 710 * (9.00×10^-1 M * 1.64×10^-2 M)
[I3-] = 1.08×10^-1 M
Now we have the equilibrium concentration of I3- after the solution is diluted to 5.50 L.
Therefore, the equilibrium ratio of [I3-] to [I2] for case a) is 5.67×10^-2 M : 9.00×10^-2 M, and for case b), it is 1.08×10^-1 M : 1.64×10^-2 M.
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.