Question

Histidine can be treated as a polyprotic acid for the purpose of making a buffer solution. The molecular weight histidine is 155.2 g/mol. You are asked to make 200 mL of a buffer solution containing 50 mM L-histidine. How much histidine do you add to the solution? Show your work. You adjust the pH of this solution to pH 5.0. Please draw the structures of the two primary histidine species at this pH. Circle the structure that is most prevalent at pH 5.0. What are the approximate concentrations of each of these two primary species in your pH 5.0 solution Please show your work. Would your solution be a good buffer against the addition of base? Why or why not?

Answer 1

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

COO

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