Buffer
a) To prepare 200 ml of 50 mM L-Histidine buffer
Histidine in buffer = 0.05 M x 0.2 L = 0.01 mmol
let x amount of base is added
pH = pKa + log(base/acid)
5 = 6.10 + log(x/0.01 - x)
x = 0.0008 - 0.08x
x = 0.00074 mol
mass of histidine to be added = (0.01 + 0.00074) x 155.2 = 1.67 g
b) at pH 5.0, the species existing in solution are shown below
circled is the major species
c) at pH 5
concentration of [histidine] = 0.00074/0.2 = 0.0037 M
concentration of [histidine HCl] = 0.00926/0.2 = 0.0463 M
Histidine can be treated as a polyprotic acid for the purpose of making a buffer solution....
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