It is desirable to prepare a solution which can serve as a buffer at a ph of 4.1

Question

It is desirable to prepare a solution which can serve as a buffer at a ph of 4.10:

a)Which of the following acids (along with their conjugate base) would be best for this purpose: Lactic Acid, Glycine Hydrochloride, Phosphoric Acid, or Propionic Acid?

b)What ratio of this chosen acid & the appropriate conjugate base would yield a buffer having a pH of exactly 4.10?

c)How would you prepare one liter of such a buffer strarting from a 1.00M solution of the acid you've selected & a 1.00M solution of Sodium Hydroxide? This buffer should be 0.050M in the selected acid.

Explanation / Answer

(a)

Following are the pKa values of the given acids :

We select the acid such that the pKa value of the acid-conjugate base pair is closest to the desired pH

Lactic acid : 3.86

Phosphoric acid : 2.15,7.2,12.35

Glycine hydrochloride : 2.4,9.8

Propionic acid : 4.87

Thus, we see, lactic acid is best suited for this buffer because its pKa is very close to the desired pH

(b)

Henderson hasselbach equation says that :

pH = pKa + log10(cb/a)

Here cb is conc of conjugate base and a is conc of acid

Putting values we get :

log10(cb/a) = pH - pKa = 4.10-3.86 = 0.24

Thus, cb/a = 100.24 = 1.73

This is the ratio of conjugate base to acid

Since we are asked the acid to conjugate base ratio, so answer is 1/1.73 = 0.57

(c)

Volume of buffer = 1L

Molarity of buffer = 0.05 M

Thus, moles of selected acid = 0.05*1 = 0.05 moles

Since the stock solution of the selected acid is 1M and we need only 0.05 moles of the acid, so the volume of stock solution needed = 0.05/1 = 0.05 L = 50 ml

Here, since we are using NaOH, so the conjugate base is sodium lactate

Let us assume we take x moles of NaOH

Due to the addition of x moles of NaOH , x moles of acid will dissociate to form conjugate base with sodium

The remaining moles ( = 0.05 - x ) remain undissociated

Thus, ratio of acid to conjugate base at equilibrium = (0.05-x)/x

As calculated above, this ratio = 0.57

Thus, (0.05-x)/x = 0.57

Thus, we get : x = 0.0318 moles

Since the stock solution of NaOH is of 1M , so volume needed = 0.0318/1 = 0.0318 L = 31.8 ml

Thus, we add 50 ml lactic acid stock solution, 31.8 ml NaOH stock solution and make the volume upto 1000 ml

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