Q19.108P - Chemistry: Molecular Nature Of Matter And Change

Question

Tris (hydroxymethyl)aminomethane [${\left({\text{HOCH}}_{\text{2}}\right)}_{\mathbf{\text{3}}}{\mathbf{\text{CNH}}}_{\mathbf{\text{2}}}\mathbf{\text{,}}$ known as TRIS] is a weak base used in biochemical experiments to make buffer solutions in the $\mathbf{pH}$ range of 7 to 9. A certain TRIS buffer has a  $\mathbf{pH}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{8}\mathbf{.}\mathbf{10}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{25}\mathbf{ }\mathbf{°}\mathbf{C}$ and a $\mathbf{pH}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{7}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{37}\mathbf{ }\mathbf{°}\mathbf{C}$. Why does the pH pH change with temperature?

Step-by-Step Solution

Answer

In order to solve this problem, we have to remember that every chemical reaction has its equilibrium constant K, which changes as the temperature change.

We know that, when we have a buffer, the Ka value is included in the Henderson-Hasselbalch equation, thus, when the Ka value changes, the pH value also changes:

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

Concept introduction

Tris(hydroxymethyl)aminomethane, sometimes known as tromethamine or THAM in medicinal settings, is an organic molecule with the formula  ${\mathbf{\left(}{\mathbf{HOCH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{3}}{\mathbf{CNH}}_{\mathbf{2}}$.

Explanation

In order to solve this problem, we have to remember that every chemical reaction has its equilibrium constant K, which changes as the temperature change.

If the reaction is exothermic:

- Temperature increases, K value decreases

- Temperature decreases,  K   value increases

 If the reaction is endothermic:

- Temperature increases,  K  value increases

- Temperature decreases,  K  value decreases

We know that, when we have a buffer, the Ka value is included in the Henderson-Hasselbalch equation, thus, when the Ka value changes, the pH value also changes.

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