Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reaction before the washing steps, and the distribution constant of the product mixture in brine is K = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume is still 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layer is 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brine wash? (See Technique 13.2, p 53.) Show work.
Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reaction before the washing steps, and the distribution constant of the product mixture in brine is K = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume is still 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layer is 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brine wash? (See Technique 13.2, p 53.) Show work.
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Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reaction
before the washing steps, and the distribution constant of the product mixture in brine is
K = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume is
still 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layer
is 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brine
wash? (See Technique 13.2, p 53.) Show work.
![---
### 13.2 Liquids
A similar washing procedure can be applied to liquid samples. Suppose that a 1.00 L aqueous solution contains 100.0 g of solute A and 10.0 g of solute B (an impurity). Suppose that solute A has a distribution coefficient \( K_A = 99.5 \), and that solute B has a distribution coefficient \( K_B = 0.100 \) for the solvent ether. That is, A is much more soluble in ether than in water, while B is more soluble in water than in ether. If the aqueous solution were extracted once with 1.00 L of ether, the ether layer would contain nearly all of the solute A (99.0 g), and a little solute B (0.909 g).
\[
K_A = 99.5 = \frac{C_{\text{ether}}}{C_{\text{water}}} = \frac{a}{(100.0 - a)} \times \frac{1.00\, \text{L}}{1.00\, \text{L}} = \frac{a}{(100.0 - a)}
\]
\[
K_B = 0.100 = \frac{C_{\text{ether}}}{C_{\text{water}}} = \frac{b}{(10.0 - b)} \times \frac{1.00\, \text{L}}{1.00\, \text{L}} = \frac{b}{(10.0 - b)}
\]
These values are obtained by solving the respective distribution coefficient equations above for a and b, respectively, where a is defined as grams of solute A dissolved in 1.00 L of ether, and b is grams of solute B dissolved in 1.00 L of ether. The layers are then separated. (In a separatory funnel, the aqueous layer is drained out, and the ether layer remains in the funnel.) Suppose that 1.00 L of pure water (wash) is then added, and the mixture is shaken vigorously until equilibrium is achieved. At this point the ether layer contains 98.0 g of solute A and only 0.0827 g of solute B. (These values were obtained by solving the following equations for \( a' \) and \( b' \), respectively, where \( a' \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feac37532-200b-4550-bb20-d71a4c33e000%2Fd9ba99d1-fdf8-44bc-9910-add6743c15cc%2Fck1dnxl_processed.png&w=3840&q=75)
Transcribed Image Text:---
### 13.2 Liquids
A similar washing procedure can be applied to liquid samples. Suppose that a 1.00 L aqueous solution contains 100.0 g of solute A and 10.0 g of solute B (an impurity). Suppose that solute A has a distribution coefficient \( K_A = 99.5 \), and that solute B has a distribution coefficient \( K_B = 0.100 \) for the solvent ether. That is, A is much more soluble in ether than in water, while B is more soluble in water than in ether. If the aqueous solution were extracted once with 1.00 L of ether, the ether layer would contain nearly all of the solute A (99.0 g), and a little solute B (0.909 g).
\[
K_A = 99.5 = \frac{C_{\text{ether}}}{C_{\text{water}}} = \frac{a}{(100.0 - a)} \times \frac{1.00\, \text{L}}{1.00\, \text{L}} = \frac{a}{(100.0 - a)}
\]
\[
K_B = 0.100 = \frac{C_{\text{ether}}}{C_{\text{water}}} = \frac{b}{(10.0 - b)} \times \frac{1.00\, \text{L}}{1.00\, \text{L}} = \frac{b}{(10.0 - b)}
\]
These values are obtained by solving the respective distribution coefficient equations above for a and b, respectively, where a is defined as grams of solute A dissolved in 1.00 L of ether, and b is grams of solute B dissolved in 1.00 L of ether. The layers are then separated. (In a separatory funnel, the aqueous layer is drained out, and the ether layer remains in the funnel.) Suppose that 1.00 L of pure water (wash) is then added, and the mixture is shaken vigorously until equilibrium is achieved. At this point the ether layer contains 98.0 g of solute A and only 0.0827 g of solute B. (These values were obtained by solving the following equations for \( a' \) and \( b' \), respectively, where \( a' \)
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