Relative to other micelles and also to isotropic bicelles, LMPG micelles were fo
ID: 485366 • Letter: R
Question
Relative to other micelles and also to isotropic bicelles, LMPG micelles were found to yield NMR spectra of superior quaility for this amyloid precursor protein. NMR samples of C99 contained 0.25 mM Amyloid precursor protein dissolved in buffer A (10% (w/v) LMPG in 100 mM imidazole (as imidazolium hydrochloride), and 10% (v/v) D2O, pH 6.5).
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)
Determine the number of moles of the acid and its conjugate base at the desired pH. Determine how many moles of weak acid and its conjugate base are present and the new pH if you were to add 10 microliters of 11.2 M solution of HCl to 5mL of buffer A.
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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moles of imidazolium salt present in 5mL =0.008 *5mL/80mL = 0.005 mol
moles of imidazole present in 5mL = 0.02 *5mL/80mL = 0.00125 mol
moles of HCl added = 10uL * 11.2M = 10*10^-6 L *11.2 M = 11.2 *10^5 mol
HCl will react with imidazolium salt to convert it to imidazole.
moles of imidazole present = 0.00125 mol + 0.000112 mol = 1.362*10^-3
moles of imidazolium salt present = 0.005- 0.000112 mol = 4.89*10^-3
pH = pka = log[imidazolium salt/imidazole]
= 6.9 + log[4.89*10^-3 /1.362*10^-3] = 7.46
New pH = 7.46