# Unusual Titration Curve Explanation

From Costanzo's Physiology book in Acid-Base Physiology chapter, this titration curve of a weak acid ($$\ce{HA}$$), in the context of discussing buffers is provided: What confused me is that in this diagram, the $$\mathrm{pH}$$ is on the $$x$$-axis rather than the $$y$$-axis. Also, why does the peek of the curve has ($$\ce{HA}$$) only and the trough has ($$\ce{A-}$$) only?

• Thanks for giving me the chance to think about it, @Poutnik. I flipped it but don't seem to understand it. As for the second question, I think that's because acids and bases occur in pairs, with an acid having a conjugate base?? Oct 18, 2020 at 13:21
• Acidic environment turns conjugate pairs toward acid side and vice versa. Oct 18, 2020 at 13:23
• Yes, but allow me to ask what does this do for our question @Poutnik ? Oct 18, 2020 at 13:25
• Do not write questions at the moment they appear, but think about them. It gives you more, if you resolve it yourself, including textbook and online resources. Only if you really failed, then ask. Oct 18, 2020 at 13:31

Acid (Brønsted–Lowry Acid) is a substance that gives away its $$\ce{H+}$$, but the environment should be able to accept that $$\ce{H+}$$. After it accepted it - the acid (which is now its conjugate base) can accept it back. There are 2 extremes:

• The environment already has plenty of $$\ce{H+}$$ already. So even if acid gives its $$\ce{H+}$$ to the environment, it immediately receives $$\ce{H+}$$ back. Hence the acid always has its $$\ce{H}$$ and you see $$\ce{HA}$$ on the picture.
• The environment is basic - it can take lots of $$\ce{H+}$$ and won't give them back to the acid. Hence acid stops being acid - it gives away $$\ce{H+}$$ and becomes $$\ce{A-}$$.

$$\mathrm{pH}$$ is a measure of how many $$\ce{H+}$$ there are in the solution. Low $$\mathrm{pH}$$ means there's plenty of $$\ce{H+}$$ (our 1st case), high $$\mathrm{pH}$$ means there's a "shortage" of $$\ce{H+}$$ (2nd case).

In the middle (where $$\ce{H+}$$ equals acid's $$\mathrm{p}K_\mathrm{a}$$ which in your case is 6.5) you'll have a case when 50% of $$\ce{H+}$$ is taken by the environment and the other 50% of acid molecules are $$\ce{HA}$$.

When you titrate you do exactly that - you change the $$\mathrm{pH}$$ of your environment which then changes how many molecules are ionized. And then at some point your indicator changes the color which means you reached some $$\mathrm{pH}$$ you were looking for.

Titration curves that I've seen have the volume of solvent added (your graph probably shouldn't be called Titration Curve?) - not $$\mathrm{pH}$$. But these are related since if you're adding a basic titrant to acidic solution you're changing its $$\mathrm{pH}$$.

• Thanks for the answer! But could you eleborate more on how does this fit on the big picture of titration, i.e how is the acid titrated this way? @Stanislav Bashkyrtsev Oct 18, 2020 at 13:43
• @Positron12, I thought I did that.. Anyway, I've added more info there. Oct 18, 2020 at 14:04
• @Positron12 Remember that $\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log{\left(\frac {\ce{[A-]}}{\ce{[HA]}}\right)}$ and the concentration ratio depends on the titration progress. Oct 18, 2020 at 14:28

For a weak acid $$\ce{HA}$$ with acid constant $$K_\mathrm{a}$$:

$$\ce{HA + H2O <=> H3O+ + A-} \tag1$$

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} \ \Rightarrow \ [\ce{H3O+}] = K_\mathrm{a} \cdot \frac{[\ce{HA}]}{[\ce{A-}]} \tag2$$

Taking $$-\log$$ of both side of equation $$(2)$$:

$$-\log [\ce{H3O+}] = -\log K_\mathrm{a} -\log \frac{[\ce{HA}]}{[\ce{A-}]}$$

Or $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A-}]}{[\ce{HA}]} \tag3$$

The equation $$(3)$$ is called Henderson-Hackleback equation (see Poutnik's comment elsewhere). According to the Le Chatelier principle, if you remove $$\ce{H3O+}$$ from the equilibrium $$(1)$$ (Note: you can do this by adding appropriate amount of strong base as a titrant, e.g., $$\ce{NaOH}$$ solution; $$\ce{H3O+ + OH- -> 2H2O}$$), more of $$\ce{HA}$$ ionizes to form $$\ce{H3O+}$$, thus, decreasing $$[\ce{HA}]$$ as a result. As a consequence, $$[\ce{A-}]$$ also increases at the same time. Therefore, $$\mathrm{pH}$$ of the solution increases according to the equation $$(3)$$.

To answer your question about "why does the peek of the curve has $$\ce{HA}$$ only and the trough has $$\ce{A-}$$ only?": Again, if $$\ce{HA}$$ is in acid medium (increasing amount of $$\ce{H3O+}$$), according to the Le Chatelier principle, the equilibrium $$(1)$$ favored the backward reaction so that more $$\ce{HA}$$ produces. At one point, if $$[\ce{H3O+}]$$ high enough (say $$\mathrm{pH} \lt 4.5$$ in your case), there won't be sufficient $$[\ce{A-}]$$, because all $$\ce{A-}$$ would be protonated to give only $$\ce{HA}$$.

During the titration (removing $$\ce{H3O+}$$), when you reached the endpoint by removing all $$\ce{H3O+}$$, you have only $$\ce{A-}$$ remaining in the solution. Yet, resulting $$\ce{A-}$$ would established following equilibrium:

$$\ce{A- + H2O <=> HA + OH-} \tag4$$

Suppose you add one extra drop of $$\ce{NaOH}$$ solution. That increase the $$[\ce{OH-}]$$ in the mixture and again, according to the Le Chatelier principle, the equilibrium $$(3)$$ favored the backward reaction so that more $$\ce{A-}$$ produces. Thus, as further addition of the titrant, only $$\ce{A-}$$ remains in the solution (of course, with $$\ce{OH-}$$, thus further increasing $$\mathrm{pH}$$).