Question

In a normal-phase partition column, a solute was found to have a retention time of $29.1$min, and an unretained sample had a retention time of $1.05$min when the mobile phase was $50\%$by volume chloroform and $50\%$n-hexane. Calculate (a) k for the solute and (b) a solvent composition that would bring k down to a value of about$10$.

Short answer

(a) Retention of the solute is $26.71$

(b) $71$percent chloroform and $29$percent hexane make up the mobile phase.

Step-by-step solution

Part(a) Step 1: Given information

Retention time of solute (t_r) $=29.1min$

Retention time of unretained sample (t_u) $=1.05min$

Mobile phase = $50\%$by volume chloroform and $50\%$n-hexane

Part(a) Step 2: Apply the formula

Partition co-efficient

$$\begin{gathered} k_1=\frac{t_r-t_m}{t_m} \\ =\frac{29.1-1.05}{1.05} \\ =26.71 \end{gathered}$$

Retention of the solute is$26.71$

Part(b) Step 1: Given information

The ratio of retention factors,

$\log \left(\frac{k_2}{k_1}\right)=\frac{p_2-p_1}{2}$.....(ii)

Polarity index of Hexane $=0.1$

Polarity index of Chloroform $=4.1$

Retention factor k₁ $=26.71$

Retention factor k₂$=10$

Part(b) Step 2: Apply the formula 

Initial Polarity index,

$$\begin{gathered} P_i=0.5P_{chloroform}+0.5P_{hexane} \\ =0.5\times 4.1+0.5\times 0.1 \\ =2.1 \end{gathered}$$

Normal Partition Column to reduce the value to $10$,

$$\begin{gathered} \log \left(\frac{10}{26.71}\right)=\frac{2.1-P_f}{2} \\ {P}_f=2.95 \end{gathered}$$

Mole fraction of chloroform = x

Mole fraction of hexane $=1-x$

Final Polarity index,

$$\begin{gathered} P_f=x{P}_{chloroform}+(1-x)P_{hexane} \\ 2.95=x\times 4.1+(1-x)0.1 \\ 2.95=4.1x+0.1-0.1x \\ x=0.71 \end{gathered}$$

Thus, the mole fraction of chloroform is $0.71$and that of hexane is $0.29$. Hence the mobile phase composition is $71\%$Chloroform and $29\%$Hexane.