In: Chemistry
Describe how you would prepare 500 mL of 0.40M glutamic acid buffer, pH 2.5, starting with solid glutamic acid in the AA2- form (MW=147.13) and 1.0M HCl. Please show work.
Glutamic buffer 500 ml of 0.40M
Volume = 500 ml = 0.500L
pH = 2.5
pka of glutamic acid = 4.2
henderson hassalbach equation
pH = pka + log [Glutamic acid/HCl]
2.5 = 4.2 + log [Glutamic acid/HCl]
2.5 - 4.2 = log [Glutamic acid/HCl]
-1.7 = log [Glutamic acid/HCl]
[Glutamic acid/HCl] = 10-1.7..................taking anntilog
[Glutamic acid/HCl] = 50.118/1
HCl = 1
Take decimal fraction
Glutamic acid = 50.118/(50.118 +1) = 0.9804
HCl = 1/0.9804 = 1.019
Calculating molarity
Glutamic acid = 0.40M x 0.9804 = 0.392 M
HCl = 1.019 x 0.40M = 0.407 M
no.of moles of HCl = 0.407 M x 0.500L = 0.203 moles
volume of HCL required = 0.203 moles/1.0 M = 0.203L ..........(1)
no. of Glutamic acid =0.392 M x 0.500L= 0.196 moles
molar mass =147.13g/mol
amount of glutamic acid = 0.196 moles x 147.13 g/mol = 28.83 g
Dissolve 28.83 g of glutamic acid in 203 ml of 1 M HCl acid make buffer of pH 2.5
given HCl = 1M
given Glutamic acid = 0.40M
no. of moles = 0.40 x 0.500L = 0.2 moles
Volume =