Question

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Answer 1

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!