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Why there is increase in the pH at the starting 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 baseenter image description here

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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
    Jan 22 at 12:05
  • $\begingroup$ What does the equation pH = pKa + log([A-]/[HA]) say to you? $\endgroup$
    – Poutnik
    Jan 22 at 12:13
  • $\begingroup$ @Poutnik . more acid will dissociate as pH increases. $\endgroup$ Jan 22 at 12:35
  • $\begingroup$ @NikhilVerma It has also the other side - increasing acid dissociation ratio by acid neutralization increases pH, $\endgroup$
    – Poutnik
    Jan 22 at 14:55

2 Answers 2

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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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Let's start from the basic formula : $$\mathrm{p}H = \mathrm{p}K_a + \log\mathrm{\frac{[A^-]}{[HA]} = \mathrm{p} K_a + {log{\frac{\mathrm{[A_o]} + n}{[HA]_o - n}}}}$$ The derivate of log$x$ is $1/x$. So when introducing $n$ mole $\ce{NaOH}$, we get a curve with a slope equal to : $$\frac{d\mathrm pH}{dn} = \mathrm{\frac{[HA]}{[A^-]}} = \frac{[\mathrm{H^+]}}{K_{a}}$$

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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