please help within five minutes

Chemistry
Tutor: None Selected Time limit: 1 Day

I-(aq) + I2(aq) I3-(aq)       K = 710

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

1.  5.00×10-2 mol of I2 is added to 1.00 L of 5.00×10-1 M KI solution.

2.The solution above is diluted to 12.00 L.


Feb 3rd, 2015

K = [I3-] / [I-][I2] = 710

                        I-(aq)         + I2(aq)          I3-(aq)

Initial               5e-1M             5e-2M          0

Change             -x                   -x               +x

Equilibrium      5e-1 -x           5e-2 -x         x


710 = x / (0.025 - 0.55x +x^2)

17.75 - 390.5x + 710x^2 = x

17.75 - 391.5x + 710x^2 =0

x = 0.04984404763809308 or x =0.5015644030661324

The first value is the only solution that works for the problem (concentraitons cannot be negative)

Ask for [I3-] / [I2] = 0.0498 / (5e-2 - 0.0498) = 0.0498/0.0002 = 249


For the second problem, just calculate the diluted concentrations and follow the same procedure.


Feb 3rd, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Feb 3rd, 2015
...
Feb 3rd, 2015
Dec 3rd, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer