Question

a. Describe the preparation of 2 liters of 0.1 M Glutamate buffer, pH 10.0. You have available monosodium glutamate (MW=169), 0.5M NaOH and distilled water. b. Momentarily distracted by having a piece of old light fixture fall on your head, you add 10ml of the 0.5 HCL to the 2L of buffer. What would be the pH now?

Answer 1

henderson hasselbalch equation

assuming the pKa of monosodium glutamate is 9.47

pOH = pKa + log([salt]/[acid])

10 = 9.47 + log([salt] / [acid])

3.38= ([salt] / [acid])

and also its given that we need to prepare 0.1 M buffer so,

[salt] + [acid] = 0.1

hence

[salt] = 0.077 M

[acid] =0.023 M

to prepare disodium glutamate

monosodium glutamate + NaOH = disodium glutamate + H₂O

meq of disodium glutamate = meq of  NaOH

0.077 X 2 = 0.5 X V

V = 770 ml

procedure

acid is monosodium glutamate

salt is disodium glutamate

to prepare monosodium glutamate 1 litre of 0.023 M , mix 3.89 g monosodium glutamate in distilled water in flask A

to prepare disodium glutamate 1 lite of 0.077 M, mix 13.01g monosodium glutamate in distilled water in flask B

mix them together and we have the required buffer.

b)

buffer resists the change of pH in the solution, so when the acid is added the buffer, it tries to neutralize the ths acid

R-COOH -----> R-COO^- + H ⁺   (where R is everything of monosodium glutamate except COOH)

Ka= ([R-COO^-] X[H⁺])/ ([R-COOH])

[H ⁺] = (10^-9.47 X 0.023)/0.077

now H+ are added that is 0.5 of 10 ml to 2 litre solution, thus the molarity of H+ become 0.00025M

and salt will combine with the H+ to nuetralize it's effect concentration of salt will decrease by 0.00025 and that of acid will incearse by 0.00025

thus the new equiblirium

Ka= ([R-COO^-] X[H⁺])/ ([R-COOH])

[H ⁺] = (10^-9.47 X 0.0232)/(0.0745)

[H ⁺]=1.05 X 10-¹⁰

pH = 9.98