The acid HA has ${\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}$.

(a) Which is the principal species, HA or ${\mathbf{A}}^{\mathbf{-}}$, at pH 6.00?

(b) Which is the principal species at pH 8.00?

(c) What is the quotient $\left[A^{-}\right]/\left[\mathrm{HA}\right]$ at pH 7.00? At pH 6.00?

Hyaluronic acid (HA), also called hyaluronan, is an anionic, nonsulfated glycosaminoglycan distributed widely throughout connective, epithelial, and neural tissues.

In this task we have an acid HA with${\mathrm{pK}}_{\mathrm{a}}$ =7.00, we need to determine which is principal species (HA or ${\mathrm{A}}^{-}$) at

1. pH =6.00M
2. pH =8.00M
3. What is the quotient $\left[{\mathrm{A}}^{-}\right]/\left[\mathrm{HA}\right]$at pH = 7.00 and pH = 6.00

Henderson-Hasselbach equation:

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-}\right]/\left[\mathrm{HA}\right]\right)$

(a)pH = 6.00

Henderson-Hasselbach equation;

$$\begin{gathered} 6.00=7.00+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ -1.00=\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ {10}^{-1.00}=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \\ 0.1=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \end{gathered}$$

(b)pH = 8.00

Henderson-Hasselbach equation;

$$\begin{gathered} 8.00=7.00+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ 1.00=\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ {10}^{1.00}=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \\ 10=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \end{gathered}$$

(c)quotient at pH = 7.00:

$$\begin{gathered} 7.00=7.00+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ 0=\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ {10}^0=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \\ 1=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \end{gathered}$$

quotient at pH = 6.00:

$$\begin{gathered} 6.00=7.00+\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ -1.00=\log \left(\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right]\right) \\ {10}^{-1.00}=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \\ 0.1=\left[{\mathrm{A}}^{-1}\right]/\left[\mathrm{HA}\right] \end{gathered}$$