Question

I2 is considerably more soluble in CCl4(l) than it is in H2O(l). At a certain temperature, the concentration of I2 in its saturated aqueous solution is 1.300×10?3 M, and the equilibrium achieved when I2 distributes itself between H2Oand CCl4 is

I2(aq)?I2(CCl4),K=85.5

A 11.0-mL sample of saturated I2(aq) is shaken with 11.0 mL of CCl4. After equilibrium is established, the two liquid layers are separated. How many milligrams of I2 will be in the aqueous layer?

Express your answer to two significant figures and include the appropriate units.

Answer: 4.2×10?2 mg If the 11.0-mL sample of aqueous layer from Part A is extracted with a second 11.0-mL portion of CCl4, how many milligrams of I2 will remain in the aqueous layer when equilibrium is reestablished? Express your answer to one significant figure and include the appropriate units. NEED THIS ONE PLEASE

If the 11.0-mL sample of saturated I2(aq) in Part A had originally been extracted with 22.0 mL of CCl4, how many milligrams of I2 will remain in the aqueous layer when equilibrium is reestablished? Express your answer to one significant figure and include the appropriate units. NEED THIS ONE PLEASE

Answer 1

To do the part B, yo only need to do the same thing of part A only that you will work with the concentration remaining from part A). In this case, According to your previous result, I calculated that and I go 4.1x10^-2 mg, and the concentration in equilibrium for both are 6.43x10^-4 M and 6.5x10^-4 (The innitial after the dillution), so, using the concentration in equilibrium:

C = 6.43x10^-4 * 11/22 = 3.215x10^-4 M

85.5 = x / 3.215x10^-4-x

85.5 * (3.215x10^-4 - x) = x

x = 0.02748825 / 86.6

x = 3.17x10^-4 M

3.215x10^-4 - 3.17x10^-4 = 4.5x10^-6 M * 0.022 = 9.9x10^-8 moles * (2*127) = 2.51x10^-2 mg

Hope this helps

Answer 2

Case-1 11ml of CClq is first added to the original aqueous sample;followed by extraction, then followed by the separati on of the residual aqueous phase; and then followed by addition of another 11ml of CCi to the resi dual aqueous phase; followed by extraction; and finally the separati on of the resultant aqueous phase for weight-determination. Weight of I2 in the aqueous phase due to the first extraction The volume of the aqueous sample (V)is11 ml The concentrati on of I2 in the aqueous sample (M)is 1.3x10 M The moles of 12 in the aqueous sample before mixing =M -3 10* L =1.3×10-3M×11ml× 1.43×10-s mol Now, the sample is shaken after the addition of 11 mL of CCl4 resulting in the transfer of I2 from the aqueous phase, and finally reaches a equilibrium value based on the distribution coefficient (K) Let x be the moles of I2 in CCl, ri ch phase at the equilibrium. So 1.43×10_-x) mol is the moles of 12 in the aqueous phase at the equilibrium The expression of X for the specified system is I2(cCI,) I2(aq) Here, I2(aq) and I2(CCl4) indicate equilibrium concentrati ons of I2 in aqueous phase and CCly respectively x mol 11mL 1.43x10-5-x) mol 11mL 85.5 .43x10--x x-1.413x 10- mol The moles of 12 in the aqueous phase at the equilibrium (n) =1.43x10-5-x= 1.7 ×10-7 mol The mass of 12 in the aqueous phase at the equilibrium-mx molar mass of 12 -1.7x10-molx2x126.904 g/mol = 4.31×10-5 g -4.3x102 mg

Weight of I2 in the aqueous phase due to the second extraction The volume of the aqueous sample ()is11ml The moles of 12 in the aqueous sample after first extraction = 1.7×10-mol Now, the sample is shaken after the addition of 11 mL of CCl4 resulting in the transfer of I2 from the aqueous phase, and finally reaches a equilibrium value based on the distributi on coefficient (K) Let y be the moles of I2 in CCl4 rich phase at the equilibrium. So, 1.7x10 -yjmolis the moles of I2 in the aqueous phase at the equilibrium. The expression of K for the specified system is I2(CCl, I2 (aq) Here, I,(aq) and I2 CCI4) indi cate equilibrium concentrati ons of I2 in aqueous phase and CCl, respectively y mol 11mL 1.7×10"-y) mol 11 mL 1.7×10"-y y 1.6803x107 mol The moles of 12 in the aqueous phase at the equilibrium (K2)-1.7 × 10-5-y = 1.97 × 10-9 mol The mass of 12 in the aqueous phase at the equilibrium-72 ×molar mass of 12 = 1.97×10-9 mol×2x126.904 g/mol = 5.00×10-7 g -5x10* mg

Case-2 When 22 mL of CClq is added at once to the original aqueous sample instead of addition of 11mL of CCl4 twice, then calculate the mass of I2 in the aqueous phase at the equilibrium as follows Weight of I2 in the aqueous phase due to the extraction Let z be the moles of in CC14 rich phase at the equilibrium. So, 1.43x10 -z) mol is the moles of L, in the aqueous phase at the equilibrium The expression of K for the specified system is, I2(CCl, I2 (aq) Here, I,(aq) and I2 CCI4) indicate equilibrium concentrations of I2 in aqueous phase and CCl, respectively 22 mL 1.43x10-5 -zmol 11 mL 85.5- 2×(1.43×10-5-r x- 1.421x10-3 mol The moles of 12 in the aqueous phase at the equilibrium (n) =1.43×10-5-2= 9×10-8 mol The mass of 12 in the aqueous phase at the equilibrium molar mass of 12 = 9×10-8 mol ×2×126.904 g/mol -2.28 x10-g -2×10-2 mg Hence extraction of I2 to CCl, in case-lis greater than that in case-2.

Answer 3

The first extraction: We can calculate the no. of moles of I2 in the first extraction by using the formula No. of mole of, before mixing Concentrati on of I2 aqueous sample -1.3000x103M Volumeofí,-(180삐×10L 1 L 10 RL 0.018 L n-(1.3000x 10-3 mol/X)x 0.013% -2.3400x10 moles After the addition of CCl4,I2 gets transferred from the aqueous layer and attains the stage of equilibrium. Let x be the no of moles of I2 in CCI Then, the no. of moles of I2 in CCl, at the stage of equilibrium is (2.3400x10 -x moles

12(aq) → 12 (CC1 4'(aq) In the above reaction, the expression of K is [I2(CCI,)] We can calculate the [I2(Ccl,)]and [I2] by using the formula, Volume of the solution is 18 mL x moles [12(CC14)」-0.018 L 2.3400x105-xmoles 0.018 L K=85.5 Substituting the above values in equation (2)

x moles 2.3400x 10 -x moles x moles 2.3400× 10% × 10%-x ) moles (85.5x 2.3400x 10%moles)-(855x x) x moles-2.3129x10 moles moles Hence, the no. of moles of I2 in CCl4 at the stage of equilibrium (2.3400x105-x moles 2.7100x107 moles Mass of I2 in CCl4 at the stage of equilibrium No. of moles x Molar mass of 12 (2.7100x1 107 moles maetes) (253 8089g oT) -6.8782 102 g In the first extraction mass of I2 in CCI4 at the stage of equilibrium is 6.8782x10 g

The second extraction Let y be the no. of moles of I2 in CCI Then, the no. of moles of I2 in CCl, at the stage of equilibrium is (2.7100x107-y moles We can calculate the [I2(CCi4)]and [I2]by using the formula, Volume of the solution is 18 mL y moles [12(CCI»)= 0.018 L 2.7100 x 10"-y)moles 0.018 L K = 85.5 Substituting the above values in equation (2) y moles 0018L 2.7100x10 -y moles () .018 . x moles 85.5= 1.4300 x103x)moles

85.5x2.7100x10'moles )-(85.5x y)-y moles y moles-2.6787×10-7 moles Hence, the no. of moles of I2 in CCl4 at the stage of equilibrium (2.7100x107-y moles -3.1300x109 moles Mass of I2 in CCl4 at the stage of equilibrium No. of moles x Molar mass of 12 = (3.1300x 10, petes)×(253.8089 gad) -(79442x10.)-107 = 7.9442 x 104 g In the second extraction mass of I2 in CCl4 at the stage of equilibrium is 9442x10 g Amount of l present in aqueous layer is 792 10