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. Use pka of glutamate 9.67.

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

[Buffer] = [Salt] + [Base]

p^H = p^K_a + log [Salt]/[Base] (Henderson-Hasselbalch equation)

10.0 = 9.67 + log [Salt]/[Base]

log [Salt]/[Base] = 10.0 - 9.67

log [Salt]/[Base] = 0.33

[Salt]/[Base] = 2.14

[Salt] = 2.14[Base]

But [Buffer] = [Salt] + [Base]

0.1 M = [Salt] + [Base]

0.1 = 2.14[Base] + [Base]

[Base] = 0.032 M

[Salt] = 0.1 - [Base]

= 0.1 - 0.032

= 0.068 M

weight of monosodium glutamate = (0.068 M x 169 x 2L)/1L

= 22.984 g

Volume of 0.5M NaOH = (0.032 M x 2000 ml)/0.5 M

= 128 ml

To prepare 2 liters of 0.1 M Glutamate buffer, pH 10.0 we have to take 22.984 g monosodium glutamate and 128 ml of 0.5M NaOH and we have to make up the solution to 2L using distilled water.

If we neglect the volume change due to addition of 10 ml HCl

Concentration of HCl in buffer solution = (0.5 M x 10 ml)/2000 ml

= 0.0025 M

0.0025 M of HCl reacts with the 0.0025 M of NaOH

new concentration of NaOH in buffer after addtion of HCl = 0.032 M - 0.0025 M

= 0.0295 M

Afer reaction with 0.0025 M of HCl 0.0025 M of NaOH will be converted into its conjugate base or Salt

  new concentration of Salt in buffer after addtion of HCl = 0.068 M + 0.0025 M

= 0.0975 M

New p^H of buffer after addtion of HCl = p^K_a + log [Salt]/[Base]

= 9.67 + log(0.0975 M/0.0295 M)

  p^H = 10.05