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Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions coming directly from the base is equal to the amount of hydronium ions due to the acid, here we have the equivalence point.

When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – andselisk Dec 3 '19 at 18:09
  • $\begingroup$ @Poutnik I saw where the mistake was; I edited now, this is what I meant $\endgroup$ – Shootforthemoon Dec 4 '19 at 16:22
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Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point.

The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount equivalent to the originally present weak acid. At this point, $[\ce{H3O+}]<[\ce{OH-}]$, so $\mathrm{pH} \gt 7$.

When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

For the approximative experimental chemical approach, it is true.

But mathematically, even this is not exactly true, as the curve is slightly asymmetric. When there is the same molar amount of hydroxide as was the acid, the solution volume is 3 times bigger. And hydroxide concentration 3 times lower.

It would be symmetric, if we constructed the graph for the constant total volume and complementing strong acid and hydroxide solution volumes, as shown in the formula below.

$$\mathrm{pH}=f\left(\frac{V_\mathrm{acid}}{V_\mathrm{acid} + V_\mathrm{hydroxide}}\right)$$

And the concentrations of both should be the same.

In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

For weak acids, the coincidence is mathematically even less true, as the curve shape before and after equivalence differs.

But for the chemical experimental point of view, the difference is still negligible, as the equivalence/inflection coincidence error is much less then experimental errors.

Both inflection and equivalence points ( approximately identical ) lay in the alkalic region.

In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

For weak acid/strong base ( and vice versa ) titrations, the neutral $\mathrm{pH}$ does not play any special role, so this point on the curve is insignificant.

Neither is there at the neutral point any inflection point, unless by chance $\mathrm{p}K_\mathrm{a}=7$. In such a case, the inflection point at the maximal acid/salt buffering capacity at the half equivalence would be at $\mathrm{p}K_\mathrm{a}=7$.

For further reference, see e.g Titration_of_a_Weak_Acid_with_a_Strong_Base

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