91Ó°ÊÓ

The principal buffering component, TRIS, has the job of keeping the pH of the buffer at a constant level, which is usually 8.0. In addition, TRIS is expected to interact with the LPS (lipopolysaccharide) in the membrane, further destabilizing it.

Let us consider the given information,

Mass of TRIS$=\text{43} \text{g}$

${\text{pK}}_{\text{3}}=5.91$

Volume of the solution $=\text{1} \text{L}$

Concentration of $\mathrm{HCl} =0.95 \mathrm{M}$

We can calculate the concentration of TRIS before any neutralization takes place:

$$\begin{gathered} \left[\mathrm{TRIS}\right]=\frac{43 \mathrm{g}}{121.36 \mathrm{g}/\mathrm{mol}×1\mathrm{L}} \\ =0.35 \mathrm{M} \end{gathered}$$

We have to calculate the concentration of TRIS after neutralization with $\text{0.095} \text{M}   \text{HCl}$.

$$\begin{gathered} \left[\mathrm{TRIS}\right]=0.35 \mathrm{M}-0.095 \mathrm{M} \\ \left[\mathrm{TRIS}\right]=0.26 \mathrm{M} \end{gathered}$$

Hence, the concentration of TRIS is $0.26 \mathrm{M}$.

The concentration of the conjugate acid formed is given below.

$\left[{\mathrm{TRISH}}^{+}\right]=0.095\mathrm{M}$

We simply use the Henderson-Hasselbalch equation now:

$\mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+{\mathrm{pK}}_{\mathrm{b}} \\ {\mathrm{pK}}_{\mathrm{a}}=\mathrm{pH}-{\mathrm{pK}}_{\mathrm{b}} \\           =14-5.91 \\           =8.09 \\ \mathrm{pH}  =  8.09+\log \frac{0.26\mathrm{M}}{0.095\mathrm{M}} \\         =8.53 \end{gathered}$$

Hence, theof the buffer is$\text{8.53}$.