Answer:
[tex]\% diss = 50\%[/tex]
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:
[tex]HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
[tex]Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}[/tex]
Thus, it is possible to find x given the pH as shown below:
[tex]x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M[/tex]
So that we can calculate the initial concentration of the acid:
[tex]\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\[/tex]
[tex][HA]_0=3.64x10^{-5}M[/tex]
Therefore, the percent dissociation turns out to be:
[tex]\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%[/tex]
Best regards!
