Describe how you would prepare exactlyof picolinate buffer, . Possible starting materials are pure picolinic acid (pyridine-2-carboxylic acid, FM 123.11), 1.0 M HCl, and . Approximately how many milliliters of the orwill be required?

• The Henderson Hasselbalch equation is a simplified formula that shows the connection between a solution's pH or pOH, pKa or pKb, and the ratio of concentrations of dissociated chemical species.
• Picolinic Acid is an endogenous metabolite of L-tryptophan (TRP) that has been shown to have a variety of neuroprotective, immunological, and anti-proliferative properties.

This work will show you how to make 100ml of 0.1M picolinate buffer with a pH of 5.50. Pure pocolinic acid (pyrid ne-2-carboxylic acid) could be used as a starting material, FM=123.11g/mol),1MHC1 and 1MNaOH.

It must first moles of picolinic acid are calculated as follows:

$$\begin{gathered} n(\mathrm{picolinic}\mathrm{acid})=\mathrm{c}\times \mathrm{V} \\ n(\mathrm{picolinic}\mathrm{acid})=0.1\mathrm{M}\times 100\mathrm{mL} \\ n(\mathrm{picolinic}\mathrm{acid})=10\mathrm{mmol} \end{gathered}$$

Consider the reaction below:

$$\begin{gathered} \mathrm{HA}+{\mathrm{OH}}^{-}\rightleftharpoons {\mathrm{A}}^{-} \\ \left[\mathrm{HA}\right]=10-\mathrm{x} \\ \left[{\mathrm{A}}^{-}\right]=\mathrm{x} \end{gathered}$$

The Henderson-Hasselbach equation should then be used to determine the value of x.

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\right[{\mathrm{A}}^{-}]/[\mathrm{HA}\left]\right) \\ 5.50=5.39+\log \left(\frac{\mathrm{x}}{10-\mathrm{x}}\right) \\ \mathrm{x}=5.63\mathrm{mmol} \end{gathered}$$

It can calculate the volume of the container from here$\mathrm{NaOH}$as a result of:

localid="1663417082743" $$\begin{gathered} V\left(\mathrm{NaOH}\right)=\mathrm{n}\times \mathrm{c} \\ V\left(\mathrm{NaOH}\right)=5.63\mathrm{mmol}\times 1\mathrm{M} \\ V\left(\mathrm{NaOH}\right)=5.63\mathrm{mL} \end{gathered}$$

Then, using the formula below, determine the mass of picolinic acid:

$$\begin{gathered} m(\mathrm{picolinic}\mathrm{acid})=\mathrm{n}\times \mathrm{M} \\ m(\mathrm{picolinic}\mathrm{acid})=10\mathrm{mmol}\times 123.11\mathrm{g}/\mathrm{mol} \\ m(\mathrm{picolinic}\mathrm{acid})=1.23\mathrm{g} \end{gathered}$$

In order to get ready 100ml of 0.1M using picolinate buffer pH=5.50 it must take the following steps:-

• Dissolve In distilled water, 1.23g of picolinic acid
• Add 5.63mL of NaOH until the pH=5.50 is reached.
• Transfer the solution to a volumetric flask and dilute to a final concentration of 100mL .