An HPLC uses a reverse-phase column with an aqueous mobile phase. The t_m for no

Question

An HPLC uses a reverse-phase column with an aqueous mobile phase. The t_m for non-retained molecules is 0.50 minutes. The equilibrium partition constant, for chlorobenzene (Cl-Benz) is 25 Cl-Benz_m double headed arrow Cl-Benz_sta And the volume ratio V_s/V_m is 0.20. Estimate the retention factor, k, for chlorobenzene in this experiment predict the retention time, t_r, for chlorobenzene in this experiment If the eluent velocity, v, is slowed down so that t_m is 1.00 minute, the peak width for chlorobenzene will be (circle one) narrower wider unchanged

Explanation / Answer

For part a) we know that the expression to use is:

K = k(Vm/Vs) then k:
k = K(Vs/Vm)

We also know that the ratio Vs/Vm = 0.20, so:
k = 0.20K

k = 0.20 * 25 = 5

b) For this part, we know that:
k = tr - tm/tm then:
k*tm = tr - tm
tr = k*tm + tm
tr = tm(k+1)

tr = 0.50(5+1)
tr = 3 min

c) finally according to the data, I think the the option b (wider) is the correct option.

Hope this helps

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