Question

To do an experiment with the peptide H-A-P-P-Y you need to make two buffers. One where the peptide has a net negative charge (Buffer A) and one where the peptide has a net positive charge (Buffer B). The pH of both buffers has to be within 2 pH units of the pI (pI=7.25), but you want the largest net charge you can have along with having the best buffer possible.

You have concentrated solutions of the following compounds available to you in the lab:
Acetic Acid (pKa = 4.8); Acetate; H3PO4 (pKa = 2); H2PO4- (pKa = 7.2); HPO42- (pKa = 12); PO43-; carbonic Acid (pKa = 6.4); bicarbonate (pKa = 10.3); carbonate; and powders of each amino acids.

Buffer A: What is the intended pH of this buffer: ____________________
Protocol to make the buffer:

Buffer B: What is the intended pH of this buffer: _____________________
Protocol to make the buffer:

Answer 1

Answer:

Buffer A: pH 5.25; 2.82 mol of acetate for every 1 mol of acetic acid

Buffer B: pH 9.25; 11.2 mol of bicarbonate for every 1 mol of carbonate

Explanation:

The pI of a protein is the pH at which it has a net charge of zero.

The protein has a positive charge below the pI and a negative charge above it.

Since pI = 7.25, let's make one buffer at pH = 5.25 and one at pH 9.25.

1. Preparation of pH 5.25 buffer

(a) Choose the conjugate pair

We should choose a buffer with pKₐ close to 5.25.

Acetic acid has pKₐ = 4.8, so let's make acetate buffer.

(b) Protocol for preparation

We can use the Henderson-Hasselbalch equation to get the acid/base ratio.

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\5.25& = & 4.8 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.45& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\2.82 & = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The base/acid ratio must be \mathbf{2.82:1} }$

To make the buffer, mix the solutions to get a ratio of 2.82 mol of acetate for every 1 mol of acetic acid.

2. Preparation of pH 9.25 buffer

(a) Choose the conjugate pair

We should choose a buffer with pKₐ close to 9.25.

Bicarbonate has pKₐ = 10.3, so let's make a pH 9.25 bicarbonate buffer.

(b) Protocol for preparation

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\9.25& = & 10.3 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\-1.05& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.08913& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\dfrac{1}{11.2}& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The acid/base ratio must be \mathbf{11.2:1} }$

To make the buffer, mix the solutions to get a ratio of 11.2 mol of bicarbonate for every 1 mol of carbonate.