Q19.118CP - Chemistry: Molecular Nature Of Matter And Change

Question

A bioengineer preparing cells for cloning bathes a small piece of rat epithelial tissue in a TRIS buffer (see Problem $\mathbf{19}\mathbf{.}\mathbf{108}$ ). The buffer is made by dissolving $\mathbf{43}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{g}$ of TRIS $\left({\text{pK}}_{\text{3}}\text{ = 5.91}\right)$ in enough  $\mathbf{0}.\mathbf{095}\mathbf{M}\mathbf{HCl}$ to make $\mathbf{\text{1.00 L}}$ of solution. What is the molarity of TRIS and the pH of the buffer?

Step-by-Step Solution

Answer

The molarity of TRIS is $0.26 \mathrm{M}$ and the of the buffer is $\text{8.53}$.

TRIS

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.

Calculating the concentration of TRIS

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:

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

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 $0.26 \mathrm{M}$.

Calculation of pH

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, the  of the buffer is $\text{8.53}$.