The distribution ratio between 3 M HCl and tri-n-butylphosphate for PdCl2 is 2.3 ¹. To calculate the percentage of PdCl2 extracted from a 25.0 mL of a 7x10^-4 M solution into 10.0 mL of tri-n-butylphosphate, we can use the following formula:

$$\text{Percent PdCl}_2 = \frac{D \times V_{\text{org}}}{C_{\text{aq}}} \times 100\%$$

where $D$ is the distribution ratio, $V_{\text{org}}$ is the volume of the organic phase, and $C_{\text{aq}}$ is the concentration of PdCl2 in the aqueous phase.

First, we need to calculate the concentration of PdCl2 in the aqueous phase. We can use the following formula:

$$C_{\text{aq}} = \frac{\text{moles of PdCl}_2}{V_{\text{aq}}}$$

where $V_{\text{aq}}$ is the volume of the aqueous phase.

We are given that the volume of the aqueous phase is 25.0 mL and the concentration of PdCl2 is 7x10^-4 M. Therefore,

$$C_{\text{aq}} = \frac{(7 \times 10^{-4} \text{ mol/L}) \times (25.0 \times 10^{-3} \text{ L})}{(25.0 \times 10^{-3} \text{ L})} = 7 \times 10^{-4} \text{ mol/L}$$

Next, we need to calculate the volume of the organic phase. We are given that it is 10.0 mL.

Now, we can calculate the percentage of PdCl2 extracted:

$$\text{Percent PdCl}_2 = \frac{(2.3) \times (10.0 \times 10^{-3} \text{ L})}{(7 \times 10^{-4} \text{ mol/L})} \times 100\% = 32.9\%$$

Therefore, approximately **32.9%** of PdCl2 will be extracted from a 25.0 mL of a 7x10^-4 M solution into 10.0 mL of tri-n-butylphosphate ¹.

Source: Conversation with Bing, 30.10.2023
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