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:

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.

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

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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml