Describe the preparation of 2.00 l of 0.100 m glycine buffer, ph 9.0, from glycine and 1.00 m naoh. what mass of glycine is required? the appropriate pka of glycine is 9.6.

pH = pKa + log ([R-]/[RH]) Where pH is the pH of the buffer, [R-] is the concentration of the basic species, and [RH] is the concentration of the acidic species. At pH 2.4, the amino group on glycine (pKa = 9.6) will be, for accounts and purposes, 100% protonated. This means our buffer will be dealing with the two ionic forms of the carboxyl group (pKa = 2.4). When pH = pKa, the two species are in equilibrium. This can be seen using the HH equation: 2.4 = 2.4 + log ([R-]/[RH]) 0 = log ([R-]/[RH]) 1 = ([R-]/[RH]) [RH] = [R-] Now we add in another equation, our conservation of mass. M = [RH] + [R-] where M is the molarity of the buffer But since [RH] = [R-]: M = 2 [RH] 0.2 = 2 [RH] And we wind up with: [RH] = [R-] = 0.1 M Now to figure out the moles of each needed, we multiply by the volume of the buffer. 0.1 M * 0.1 L = 0.01 mol This shows that to make 100 ml of 0.2 M glycine buffer, we'll need 0.01 mol of each species. 0.01 mol of 0.5 M HCl: 0.5 mol HCl / 1 L = 0.01 mol / v solve for v v/1 = 0.01 / 0.5 ==> v = 0.02 L or 20 mL weight of glycine: MW: 75.07 g/mol 0.01 mol glycine * (75.07g glycine / 1 mol) = 0.75 g glycine And there's your answer To make this buffer you would add 0.75g glycine to 20 mL of 0.5 M HCl and fill with water until a 100mL volume was achieved.

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