Q41P - Organic Chemistry (Mcmurry)

Question

In humans, the final product of purine degradation from DNA is uric acid, $\mathbf{pKa}=\mathbf{5}.\mathbf{61}$, which is excreted in the urine. What is the percentdissociation of uric acid in urine at a typical $\mathbf{pH}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}$? Why do you think uric acid is acidic even though it does not have a ${\mathbf{CO}}_2\mathbf{H}$ 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 $p{K}_a=5,61$. On representing and substituting Henderson-Hasselbalch equation,

$\begin{array}{l}\log \frac{\left[HA\right]}{\left[A^{-}\right]}=p{K}_a-pH or \log \frac{\left[A^{-}\right]}{\left[HA\right]}=pH-p{K}_a\\ \log \frac{\left[A^{-}\right]}{\left[HA\right]}=6-5.61\\ \log \frac{\left[A^{-}\right]}{\left[HA\right]}=0.39\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log of -0.39\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=2.45\\ \left[A^{-}\right]=2.45\left[HA\right]---\left(1\right)\end{array}$

Step 2: Finding percent dissociation of uric acid

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

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

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.