Q42E - Organic Chemistry (Mcmurry)

Question

In humans, the final product of purine degradation from DNA is uric acid, pKa = 5.61, which is excreted in the urine. What is the percent dissociation of uric acid in urine at a typical pH = 6.0? Why do you think uric acid is acidic even though it does not have a CO₂H group?

Step-by-Step Solution

Answer

The dissociation percentage of uric acid is 71.01 %. The uric acid is acid due to its conjugate base got by proton loss stabilized by resonance.

Step 1: Finding [A - ] by representing HA and A -

Given that pH=6 and pK_a=5.61. On representing and substituting Henderson-Hasselbalch equation,

$$\begin{gathered} \log \frac{\left[\mathrm{HA}\right]}{\left[{\mathrm{A}}^{-}\right]}={\mathrm{pK}}_{\mathrm{a}}-\mathrm{pH} \mathrm{or} \log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=\mathrm{pH}-{\mathrm{pK}}_{\mathrm{a}} \\ \log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=6-5.61 \\ \log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=0.39 \\ \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=\mathrm{antilog} \mathrm{of}-0.39 \\ \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=2.45 \\ \left[{\mathrm{A}}^{-}\right]=2.45\left[\mathrm{HA}\right]---\left(1\right) \end{gathered}$$

Step 2: Finding percent dissociation of uric acid

There is a presence of both ionized uric acid at equilibrium and also un-ionized forms.

$$\begin{gathered} \left[\mathrm{HA}\right]+\left[\mathrm{A}-\right]=100\%---\left(2\right) \\ \left[\mathrm{HA}\right]+2.45\left[\mathrm{HA}\right]=100\% \\ 3.45\left[\mathrm{HA}\right]=100\% \\ \left[\mathrm{HA}\right]=\frac{100}{3.45} \\ \left[\mathrm{HA}\right]=28.99\% \\ \left[{\mathrm{A}}^{-}\right]=100-28.99 \\ \left[{\mathrm{A}}^{-}\right]=71.01\% \end{gathered}$$

The uric acid contains 3 hydrogen atoms attached to atoms of N. The proton removal will yield stabilization of resonance as conjugate base. The uric acid is acid nature.