# Percent Dissociated from Titration Curve

Question 818 references the titration curve. Answer is A because $$\ce{H+}$$ conc $$= 10^{-4}$$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $$0.1 - 0.0001$$, which is about $$0.1$$. So then its $$\frac{0.0001}{0.1}\times 100 = 0.1\%$$.

Where in the world are they getting the original concentration? how did they get $$0.1$$ as the concentration of undissociated acid?

I'm not sure I follow your logic. For a monobasic acid S $$(\ce{HA})$$ dissociation degree $$α$$ is

$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$

where $$c_\mathrm{a}$$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $$V_\mathrm{b}$$:

$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies c_\mathrm{a} = \frac{c_\mathrm{b}V_\mathrm{b}}{V_\mathrm{a}}$$

Finally, taking $$V_\mathrm{a} = V_\mathrm{b} = \pu{50 mL}$$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $$[\ce{H+}] \approx \pu{1e-4 mol L-1}$$:

$$α = \frac{[\ce{H+}]V_\mathrm{a}}{c_\mathrm{b}V_\mathrm{b}} = \frac{\pu{1e-4 mol L-1}\cdot\pu{50 mL}}{\pu{0.1 mol L-1}\cdot\pu{50 mL}} = \pu{1e-3}~\text{or}~0.1\%$$

• Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value – Jon Apr 1 '19 at 15:00