Question

Iodine is sparingly soluble in pure water. However, it does 'dissolve' in solutions containing excess iodide ion because of the following reaction: 1(aq) + 12(aq) = 13(aq) K = 710. For each of the following cases calculate the equilibrium ratio of [13] to [12]. 5.00x10-2 mol of I2 is added to 1.00 L of 5.00x10-1 M KI solution. 3.20x102 You are correct. Previous Tries The solution above is diluted to 12.50 L. 7.09 Submit Answer You have entered that answer before Incorrect. Tries 3/99 Previous Tries

Answer 1

Initially 17.) - 12 5,00 x 10 mol 5.00x10-2M 100L HOOL 1. KI] = ·5.00X10"'M Is - + - I I2 I 5X107 x 102 с -1 - E 2X10 - 3x1024 Equilibrium constant - FH The 710 = (5xolu ») C5*10^-^) ) 710x (0.5-w) (005-) = x 2 m 910 (0:02 5-0.591 +*") = * 17.75 -390.5.5 X 6737101" an. = Scanned with CamScanner

71009~391151 +17.75 = 0 its is a quadratic Equation Solving giver n=0.0498 ad or it will lead N=0.501 , nato The Value n=0.501 is neglected as it to negative concentration for iodine, Lune Hence x = 0.0898 At Equilibrium [13] = x Lo Gooxon = 0.0698 , 205 m. Ov =249 @ The Solution is diluted to 18.50L Tibully Initially 5x10mol = 0.05 19,50.- 19.50 0:004 M.OP 6 - éross E Scanned with CamScanner

(11) - SKI] = - 5X10'M X 1.0L = 0.04.14 19.50 I . Iz Iz 0.04 0.0040 -* :-* +7 0-04-X 0.004-x x I Ć E K z_653 [3"] [43] 70 = 10.04-) (0,004-1) - 710 (0.04-u) (0.004-X) -X = 7 0.1136-31.9477% =) Fox-32.242 + 0.1136=0] on solving X = 0.0415, CS Scanned N = 0.00384. CamScanner

Therefore, support y = 0:00384. at Equilibrium [55] - {In] 9 = 0.00384 0.05-0.00384 = 0.083 iki pichon - Wool of Scanned with CamScanner