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Describe how you would prepare 500 mL of 0.40M glutamic acid buffer, pH 2.5, starting with...

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.

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Expert Solution

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 =

queen_honey_blossom answered 1 year ago

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