Question

For a TRIS buffer with a \( \mathrm{pKa} \) value of 8.30 and a solution with a pH of 6.50, what is the ratio of [TRIS]/[TRIS-H+]?

Answer 1

0.0158489

Step-by-step explanation:

Okay, so the ratio of [TRIS]/[TRIS-H+] in a buffer solution can be calculated with the Henderson-Hasselbalch equation. This equation is as follows:

\[

\mathrm{pH} = \mathrm{pKa} + \log_{10} \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right)

\]

where

- \(\mathrm{pH}\) is the pH of the solution

- \(\mathrm{pKa}\) is the acid dissociation constant

- \([\mathrm{A}^-]\) is the concentration of the base (in this case, TRIS)

- \([\mathrm{HA}]\) is the concentration of the acid (in this case, TRIS-H+)

We want to find the ratio \(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\), which is equivalent to \(\frac{[\mathrm{TRIS}]}{[\mathrm{TRIS-H+}]}\), so we can rearrange the Henderson-Hasselbalch equation as follows:

\[

\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 10^{\mathrm{(pH - pKa)}}

\]

Now, we can substitute the given values into the equation:

\[

\frac{[\mathrm{TRIS}]}{[\mathrm{TRIS-H+}]} = 10^{(6.50 - 8.30)} = 10^{-1.80}

\]

Hence, the ratio of [TRIS]/[TRIS-H+] is \(10^{-1.80}\) or approximately 0.0158489.

Hope this helps! :)