Q40P - Organic Chemistry (Mcmurry)

Question

Thioglycolic acid,  ${\mathbf{HSCH}}_2{\mathbf{CO}}_2\mathbf{H}$, a substance used in depilatory agents
 (hair removers) has $\mathbf{pKa}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{42}$. What is the percent dissociation of
 thioglycolic acid in a buffer solution at $\mathbf{pH}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}$?

Step-by-Step Solution

Answer

The dissociation percentage of thioglycolic acid is $\mathbf{-}\mathbf{27}\mathbf{.}\mathbf{55}\mathbf{\%}$. 

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

Given that $pH=3$ and $p{K}_a=3.42$. On representing and substituting Henderson-Hasselbalch equation,

$\begin{array}{l}\log \frac{\left[HA\right]}{\left[A^{-}\right]}=p{K}_a-pH or \log \frac{\left[A^{-}\right]}{\left[HA\right]}=pH-p{K}_a\\ \log \frac{\left[A^{-}\right]}{\left[HA\right]}=3-3.42\\ \log \frac{\left[A^{-}\right]}{\left[HA\right]}=0.42\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=anti\log of -0.42\\ \frac{\left[A^{-}\right]}{\left[HA\right]}=0.38\\ \left[A^{-}\right]=0.38\left[HA\right]---\left(1\right)\end{array}$

Step 2: Finding percent dissociation of thioglycollic acid

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

$\begin{array}{c}\left[HA\right]+\left[A^{-}\right]=100\%---\left(2\right)\\ \left[HA\right]+0.38\left[HA\right]=100\%\\ 1.38\left[HA\right]=100\%\\ \left[HA\right]=\frac{100}{1.38}\\ \left[HA\right]=27.55\%\end{array}$