Figure \(15-1\) (page 656 ) shows that \(I_{2}\) is considerably more soluble in \(\mathrm{CCl}_{4}(1)\) than it is in \(\mathrm{H}_{2} \mathrm{O}(1) .\) The concentration of \(I_{2}\) in its saturated aqueous solution is \(1.33 \times 10^{-3} \mathrm{M},\) and the equilibrium achieved when \(\bar{I}_{2}\) distributes itself between \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CCl}_{4}\) is $$\mathrm{I}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) \quad K_{\mathrm{c}}=85.5$$ (a) \(\mathrm{A} 10.0 \mathrm{mL}\) sample of saturated \(\mathrm{I}_{2}(\mathrm{aq})\) is shaken with \(10.0 \mathrm{mL} \mathrm{CCl}_{4} .\) After equilibrium is established, the two liquid layers are separated. How many milligrams of \(I_{2}\) will be in the aqueous layer? (b) If the \(10.0 \mathrm{mL}\) of aqueous layer from part (a) is extracted with a second \(10.0 \mathrm{mL}\) portion of \(\mathrm{CCl}_{4}\) how many milligrams of \(\mathrm{I}_{2}\) will remain in the aqueous layer when equilibrium is reestablished? (c) If the 10.0 mL sample of saturated \(I_{2}(\) aq) in part (a) had originally been extracted with \(20.0 \mathrm{mL} \mathrm{CCl}_{4}\) would the mass of \(I_{2}\) remaining in the aqueous layer have been less than, equal to, or greater than that in part (b)? Explain.

Step by step solution

01

Calculate initial mass of \(I_{2}\) in \(H_{2}O\)

The initial concentration of \(I_{2}\) given is \(1.33 \times 10^{-3} \, M\). This is the amount of \(I_{2}\) in a \(1.0 \, L\) or \(1000.0 \, mL\) of water. If we have \(10.0 \, mL\) of the solution, the initial moles of \(I_{2}\) will be \(\frac{10.0}{1000.0} \times 1.33 \times 10^{-3} = 1.33 \times 10^{-5} \, mol\). Since the molar mass of \(I_{2}\) is approximately \(253.8 \, g/mol\), the initial mass of \(I_{2}\) in the solution is \(1.33 \times 10^{-5} \, mol \times 253.8 \, g/mol \approx 3.38 \, mg\).

02

Calculate quantity of \(I_{2}\) transferred to \(CCl_{4}\) layer

The equilibrium constant \(K_{c}\) is 85.5, which means the reaction favors the right side; implying that more \(I_{2}\) will dissolve in \(CCl_{4}\) than in \(H_{2}O\). The equilibrium equation is \([I_{2}(CCl_{4})] = K_{c} \times [I_{2}(H_{2}O)]\). As the volumes of \(H_{2}O\) and \(CCl_{4}\) are the same, the equilibrium moles of \(I_{2}\) in \(CCl_{4}\) is \(85.5 \times [I_{2}(H_{2}O)]\). From the equilibrium equation, we can calculate the mass of \(I_{2}\) transferred to \(CCl_{4}\), which is \(3.38 \, mg - 3.38 \, mg / (85.5 + 1) \approx 3.20 \, mg\). Thus, \(0.18 \, mg\) of \(I_{2}\) remains in \(H_{2}O\). This is the answer to part (a).

03

Adjust for a second extraction

After the second \(10.0 \, mL\) portion of \(CCl_{4}\) is introduced, the process repeats itself. We apply the same equilibrium calculation to the remaining mass of \(I_{2}\) in the \(H_{2}O\). So, the mass of \(I_{2}\) remaining in \(H_{2}O\) after the second extraction is \(0.18 \, mg - 0.18 \, mg / (85.5 + 1) \approx 0.17 \, mg\). This is the answer to part (b).

04

Compare to a single extraction with \(20.0 \, mL\) \(CCl_{4}\)

In part (c), instead of two separate \(10.0 \, mL\) extractions of \(CCl_{4}\), if \(20.0 \, mL\) was used all at once, the mass of \(I_{2}\) left in \(H_{2}O\) would have been less. This is due to the increased quantity of \(CCl_{4}\) which will absorb more iodine based on the equilibrium constant. That's because the volume of \(CCl_{4}\) is a factor in the equilibrium calculation, so using a larger volume once would result in more \(I_{2}\) being extracted.

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