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Why is there an increase in the pH at the start of a titration between a weak acid and a strong base even though there is acid already present to neutralize it? My intuition says that there should not be even a slight increase in pH until all of the weak acid is dissociated and neutralized by the base

titration curve

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    $\begingroup$ The StackExchange network expects that you have thoroughly searched and thought about the topic and have provided explicit compact summary of partial answers/ideas/thoughts you have got until now. That will prevent responders to tell you what you already know or what can be easily found. Effort not shown may be considered as effort not done and such a question may get closed. How do I ask a good question?. $\endgroup$
    – Poutnik
    Commented Jan 22, 2022 at 12:05
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    $\begingroup$ What does the equation pH = pKa + log([A-]/[HA]) say to you? $\endgroup$
    – Poutnik
    Commented Jan 22, 2022 at 12:13
  • $\begingroup$ @Poutnik . more acid will dissociate as pH increases. $\endgroup$
    – Nikhil Verma
    Commented Jan 22, 2022 at 12:35
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    $\begingroup$ @NikhilVerma It has also the other side - increasing acid dissociation ratio by acid neutralization increases pH, $\endgroup$
    – Poutnik
    Commented Jan 22, 2022 at 14:55

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Because the weak acid, in this case, is not all that weak (the curve looks like one that would be made with acetic acid). There is a small supply of dissociated, aqueous hydrogen ions with which the base reacts first, before establishing its buffer equilibrium with the weak-acid molecule and its anion. The initial increase in pH then corresponds to exhausting this hydrogen-ion supply.

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  • $\begingroup$ The curve shown in the image is a classical example of a weak acid titration you can find a lot of them in literature. Why do you say that the reason if that the acid is not so weak? Can you find a curve of a weak acid and a strong base titration where the pH at the beginning doesn't rise? $\endgroup$
    – G M
    Commented Jul 25, 2022 at 6:11
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Whenever the addition of small amounts of strong base changes the ratio of weak acid to weak conjugate bases a lot, the slope will be high. This is the case at the beginning (when the concentration of weak base is low) and near the equivalence point (when the concentration of weak acid is low). When the ratio of weak acid to weak base is 1:1 (at the half equivalence point), buffering is optimal and the slope is minimal. All of this ignores the volume changes when adding the strong base.

In effect, the strong base reacts mostly with hydronium ions in the beginning (steep slope), mostly with the weak acid when nearing the half-equivalence point (shallow slope), and with nothing past the equivalence point (steep slope until you reach extremely basic pH).

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The $\mathrm{pH}$ is a dynamic equilibrium linked with the presence of both $\ce{OH-}$ and $\ce{H3O+}$ ions. $$ \mathrm{pH} + \mathrm{pOH} = 14 $$ When you are removing $\ce{H3O+}$ the equilibrium shifts. So you will have an increase in the $\mathrm{pH}$. Hence, you can't avoid perturbing this equilibrium when adding one of the two species, of course, the highest deviation from the equilibrium will be after that you consume all the $\ce{H3O+}$ that's why we use a $\mathrm{p}$ function. In fact, that's why we use a $\log$ function, the variation is not so high seen from the point of view of concentration on a linear scale.

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  • $\begingroup$ The question can be quite reduced to Why z changes when x/y changes (and vice versa) in z = a.(x/y)? $\endgroup$
    – Poutnik
    Commented Jul 26, 2022 at 8:31
  • $\begingroup$ @Poutnik probably the question is not well posed, there is an increase of the pH in every titration between an acid (week or strong) and a base. So the answer can not be because is not a weak acid as it seems from the other answers. $\endgroup$
    – G M
    Commented Jul 27, 2022 at 9:32
  • $\begingroup$ Sure. The main difference is concave initial curvature for weak acids versus convex one for strong acids. $\endgroup$
    – Poutnik
    Commented Jul 27, 2022 at 10:18
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Let's start from the basic formula:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]} = \mathrm{p}K_\mathrm{a} + \log\frac{\mathrm{[\ce{A}]_0} + n}{[\ce{HA}]_0 - n}\tag{1}$$

The derivative of $\log x$ is $1/x$. So when introducing $n$ mol $\ce{NaOH}$, we get a curve with a slope equal to: $$\frac{\mathrm {dpH}}{\mathrm dn} = \frac{[\ce{HA}]}{[\ce{A^-}]} = \frac{[\ce{H^+]}}{K_\mathrm{a}}\tag{2}$$

At the beginning of the titration, when $n$ is small, the acidic concentration $[\ce{H+}]$ is high. So the slope of the curve is steep. When $n$ grows bigger, $[\ce{H+}]$ becomes smaller. As a consequence, the slope of the curve decreases.

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  • $\begingroup$ The derivative is not quite complete, you are missing the inner term (and the n in numerator and denominator sums or differences. According to Wolfram Alpha, d/dx(log((a + x)/(b - x))) = (a + b)/((a + x) (b - x)). This assumes natural log. $\endgroup$
    – Karsten ♦
    Commented Jul 22, 2022 at 4:44
  • $\begingroup$ @Karsten Theis. Your derivation is correct. But it is not necessary. The question was not to get a numerical value for the slope. It was only to understand qualitatively why the initial slope pH vs n(NaOH) decreases at the beginning of the titration. So my derivation is valid. And if the whole calculation is done with the initial values (following your method), the final value of $\frac{d\mathrm pH}{dn}$ is the same as mine. $\endgroup$
    – Maurice
    Commented Jul 22, 2022 at 9:01

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