# Question 32cf0

May 1, 2015

Warning. This is a long answer because the question has many parts.

A solute will distribute itself between two immiscible liquids according to its solubilities in those liquids.

If one liquid is an organic solvent and the other is water, we can write

${K}_{\text{d" = c_"w"/c_"o}}$

where ${c}_{\text{w}}$ is the equilibrium concentration in water, ${c}_{\text{o}}$ is the concentration in the organic solvent, and ${K}_{\text{d}}$ is a constant called the distribution coefficient or the partition coefficient.

Experimental

(Bottle 1) Shake 40.0 mL of 1.00 mol/L acetic acid with 20 mL of butan-1-ol. Then titrate 10.0 mL samples of each layer with 0.500 mol/L $\text{NaOH}$.

Note: The numbers below are made-up examples.

Calculations

Moles of $\text{HA}$ in original sample

Your original solution of $\text{HA}$ contained

40.0 cancel("mL HA") × "1.00 mmol HA"/(1 cancel("mL HA")) = "40.0 mmol HA"

Moles of $\text{HA}$ in water layer

Volume of $\text{NaOH}$ to titrate water layer = 16.0 mL

$\text{Moles of HA" = 16.0 cancel("mL NaOH") × (0.500 cancel("mmol NaOH"))/(1 cancel("mL NaOH"))× "1 mmol HA"/(1 cancel("mmol NaOH")) = "8.00 mmol HA}$

But you used only 10.0 mL of the 40.0 mL water layer.

The water layer must have contained

40.0 cancel("mL") × "8.00 mmol HA"/(10.0 cancel("mL")) = "32.0 mmol HA"

(i) Volume of NaOH to titrate organic layer

The original acetic acid solution contained 40.0 mmol $\text{HA}$.

After shaking with butan-1-ol, the water layer contained 32.0 mmol $\text{HA}$.

So the organic layer contained

$\text{(40.0 – 32.0) mmol HA" = "8.0 mmol HA}$

But you titrated only 10.0 mL of the 20.0 mL layer or

10.0 cancel("mL") × "8.0 mmoL HA"/(20.0 cancel("mL")) = "4.0 mmol HA"

The volume of $\text{NaOH}$ required was

4.0 cancel("mmol HA") × (1 cancel("mmol NaOH"))/(1 cancel("mmol HA")) × "1 mL NaOH"/(0.500 cancel("mmol NaOH")) = "8.0 mL NaOH"

(ii) Distribution coefficient

The water layer contained 32.0 mmol $\text{HA}$, and the organic layer contained 8.0 mmol $\text{HA}$.

K_"d" = c_"w"/c_"o" = (32.0 cancel("mmol"))/(8.0 cancel("mmol")) = 4.0#

(iii) Bottle 2

I leave you to repeat the above calculations for Bottle 2, which contained 20.0 mL of butan-1-ol, 30.0 mL of $\text{HA}$, and 10.0 mL of water (still 20.0 mL organic + 40.0 mL aqueous).

(iv) Assumptions

• The solvents are immiscible
• The temperature is constant
• The distribution is at its equilibrium value

(v) Which indicator?

You are titrating a weak acid with a strong base, so the pH at the equivalence point will be basic.

Use phenolphthalein (pH 9).

(vi) pH change in aqueous layer?

The concentration of the original HA was 1.00 mol/L.

After shaking with butan-1-ol, the concentration was less than 1.00 mol/L.

The solution was less acidic (more basic), so the pH increased (perhaps from pH 3 to pH 4).