(a) Make a graph showing retention times of peaks 6, 7, and 8 in Figure 25-12 as a function of %acetonitrile (%B). Predict the retention time of peak 8 at 45% B.

(b) Linear-solvent-strength model: In Figure 25-12, t_m = 2.7 min. Compute k for peaks 6, 7, and 8 as a function of %B. Make a graph of log k versus Φ, where Φ= %B/100. Find the equation of a straight line through a suitable linear range for peak 8. The slope is -S and the intercept is log k_w. From the line, predict t_r for peak 8 at 45% B and compare your answer with (a).

(c) Gradient elution: A linear eluent gradient from 40 to 80% acetonitrile over 30 min is performed on the column in Figure 25-12. Assuming a dwell volume of 0 mL, use your data from (b) to plot the retention factor of peaks 6 and 8 during the gradient. What are the general characteristics of the plot?

(d) Why are the peaks in a gradient separation sharp?

(a)

From figure 25-12 we observed the retention time of peak 6,7 and 8 and plotted the values as a function of %B(acrylonitrile). The table used to plot the graph is shown above.

The graph for peak 6, 7 and 8 obtained from the table is plotted below. It is asked topredict the retention time of peak 8 at 45% B. As per assumption at 45% B peak 8 may be eluted halfway between 40% B and 50% B which is around 45 min.But as per the curve fitting it suggests 38.6 min is a more accurate time. The value obtained is shown by dotted line here.

Fig1: graph showing retention times of peaks 6, 7, and 8

(b)

In Figure 25-12, t_m = 2.7 min. We need to find k for peaks 6, 7, and 8 as a function of %B and need to plot a graph of log k versus Φ, where Φ= %B/100. The data regarding log k value is shown in the table below.

The retention factor (k) can be calculated using the following equation

$k=\frac{t_r-t_m}{t_m}$

t_r = Retention time

The graph using the above table is plotted below:

Fig2: graph showing log k vs solvent composition of peaks 6, 7, and 8

Now we need to find the equation of a straight line through a suitable linear range for peak 8.

From the trend line of peak 8 the linear equation obtained $\log k=-3.2682\varphi +2.6364$

For 45 % B lets plug inΦ=0.45 in the linear equation

localid="1660827850931" $\begin{array}{rcl}\log k& =& -3.2682\times 0.45+2.6364\\ \log k& =& 1.16571\\ k& =& 14.65\\ k& =& \frac{t_r-t_m}{t_m}\\ \frac{t_r-t_m}{t_m}& =& 14.65\\ \frac{t_r-2.7}{2.7}& =& 14.65\\ t_r& =& 42.255\end{array}$

Therefore, atΦ=0.45, k=14.65 and t_r=42.255 min. The value obtained from graph in part (a) is 38.6 min which is slight less than that obtained in part (b)

A linear eluent gradient from 40 to 80% acetonitrile over 30 min is performed on the column in Figure 25-12. Assuming a dwell volume of 0 mL, data from (b) was used to plot the retention factor of peaks 6 and 8 during the gradient. From the graph it was obtained that for both the peaks retention factor is high at first and then decreases gradually following exponential behavior until the compounds are almost unretained at the end of the gradient. The graph obtained with exponential trend line and R² value is shown below.

Fig 3: graph showing retention times of peaks 6, 7, and 8 for gradient elution

(d) In gradient elution the zones are sharpened due to the compression caused by the different retention factors at the leading and trailing part of the peak. By the time a compound is eluted, it is almost unretained and has the peak width of a weakly retained compound.