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Figures 11-1a and 11-1b

SO, in this picture, two different titration curves are shown: one for a weak acid, one for a strong acid. I'm confused as to why for a strong acid, if you vary the concentration of the acid, (and the concentration of the acid and titrant are the same) the pH lines are spread apart below the equivalence point, but for a weak acid, where the pHs are initially slightly different, the lines converge to a common path

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    $\begingroup$ It is due to the fact that at half equivalence point, the pH of the solution is equal to the pKa value of the weak acid. And this pH does not depend on the initial concentration of the acid. You should take into account something that does not appear on your diagrams. The concentration of the strong base (used on the abscissa) is not given ! They are not the same in the three titrations. They are equal to the concentration of the weak or strong acid. $\endgroup$ – Maurice May 17 at 9:31
  • $\begingroup$ @Maurice I agree. But you should turn your comment into an answer :) $\endgroup$ – William R. Ebenezer May 17 at 11:24
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    $\begingroup$ Consider the situation exactly halfway to the equivalence point. For the weak acid cases, the pH equals the pKa in all three cases: this is the center of the buffer region. For the strong acid cases, the added NaOH was completely neutralized, so the hydrogen ion concentrations decrease by a factor of two (because of the neutralization) and also by the dilution caused by adding the NaOH solution. So the pH values halfway to the equivalence are still separated by a pH unit for values in your example. $\endgroup$ – Ed V May 17 at 13:29
  • $\begingroup$ Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$ – Ed V May 26 at 1:39
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To understand what is happening with the two sets of titration curves, start by considering the titration of acetic acid ("HOAc" for short) with sodium hydroxide solution. The titration curves are shown below:

HOAc titration curves

In the figure, the mutual concentrations are simply the concentrations of the reactants. Thus, 0.1 M HOAc is titrated with 0.1 M NaOH, 0.01 M HOAc is titrated with 0.01 M NaOH, and so on.

Acetic acid is a monoprotic weak acid, with $\mathrm{p}K_\mathrm{a}=4.756$. The titration curves for the three highest mutual concentrations almost coincide at the half-equivalence points, i.e, at the center of the buffer region. This is as expected from the Henderson-Hasselbalch equation, which ignores the autoionization of water. However, the three lower concentrations demonstrate that the titration curves do not coincide at the half-equivalence points: as the concentrations decrease, the pH values at the half-equivalence pints move toward the assumed pH of 7 for pure water.

As the concentrations are lowered, the effect of water's auto-ionization can no longer be neglected and the Henderson-Hasselbalch approximation is no longer viable. The calculated pH values at the half-equivalence points are shown in the second column of the table below:

Table of pHs

Note that for 0.1 M, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $, as expected. Even at 0.001 M, the approximation is not bad. However, as concentrations continue to decline, the pH at the half-equivalence points simply move even closer to that assumed for pure water.

Now consider the titration of hydrochloric acid with sodium hydroxide solution. The titration curves are shown below:

Titration of HCl

As shown in the figure and in the third column of the table, the pH values at the half-equivalence points are one pH unit apart until the concentration is down around $\mathrm{1 \mu M}$. Then, as concentrations continue to decline, the pH at the half-equivalence points must invariably move toward that of pure water.

For hydrogen ion concentrations high enough to neglect the auto-ionization of water, the pH at the half-equivalence point is easy to calculate. First, divide the initial hydrogen ion concentration by 2, to account for the fact that half of the number of hydrogen ions, from the 50 mL of starting HCl solution, were neutralized by addition of 25 mL of equimolar NaOH solution. Then multiply the hydrogen ion concentration by 2/3, because the solution volume increased from 50 mL to 75 mL. The net result is that the hydrogen ion concentration, at the half-equivalence point, is 1/3 of the initial value. Then take the negative logarithm base 10, as usual:

$$\mathrm{pH} = -\log{({\ce{[H+] \times (1/2) \times (2/3)) = - \log{(\ce{[H+]/3)}}}} }$$

If $\ce{[H^+]}$ = 0.1 M, then pH = 1.477. N.B. Significant figures were ignored in this answer, so 0.1 M is treated as exact, etc.

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It is due to the fact that at half equivalence point, the pH of the solution is equal to the $\ce{pK_a}$ value of the weak acid. And this pH does not depend on the initial concentration of the acid. You should take into account something that does not appear on your diagrams. The concentration of the strong base (used on the abscissa) is not given ! They are not the same in the three titrations. They are equal to the concentration of the weak or strong acid.

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