LMPG is a phospholipid-like detergent. Relative to other micelles and also to is
ID: 485113 • Letter: L
Question
LMPG is a phospholipid-like detergent. Relative to other micelles and also to isotropic bicelles, LMPG micelles were found to yield NMR spectra of superior quaility for this protein. NMR samples of C99 contained 0.25 mM protein dissolved in buffer A (10% (w/v) LMPG in 100 mM imidazole (as imidazolium hydrochloride), and 10% (v/v) D2O, pH 6.5). NMR data were collected at 45 dergrees celsius.
Explain how you would prepare 80 mL of buffer A at the target pH. Look up molecular weights and provide in grams, how much of each component needs to be added, or if added from a concentrated stock, the required volume. What will the pH of the solution be if you initially dissolve the buffer in 1/2 the target volume? (imidazolium, pka = 6.90)
Explanation / Answer
pH of a buffer is dictated by Hinderson Hasselbalch equation.
pH = pKa +log[imidazolium salt/imidazole]
6.5 = 6.9 + log [imidazolium salt/imidazole]
[imidazolium salt/imidazole] = 0.398
According to given problem [inidazolium hydrochloride] = 100mM = 0.1M
Moles of Imidazolium salt = 0.1 M *0.08 L = 0.008 moles
Mass of imidazolium salt = 0.008 moles *104.58 g/mol =0.837 g
Inidazole = 0.1M/0.398 = 0.251 M
moles of imidazole = 0.251 M * 0.08 L = 0.02 moles
Mass of imidazole = 0.02 moles *68.077 g/mol = 1.37 gm
Hence, to prepare this buffer 1.37gm imidazole and 0.837 gm imidazolium salt has to be added to 80mL water.
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Ph of the solution if the same amount is dissolved in 40mL water
moles of imidazole = 0.02 mol
Molarity = 0.02/0.04 L = 0.5 M
moles of imidazolium salt = 0.008
Molarity = 0.008/0.04 = 0.2 M
pH = pka + log[imidazolium salt/imidazole]
= 6.9 + log[0.2/0.5] = 6.5
pH will be same as molarity of both imidazole and imidazolium salt will change to the same extent.
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