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You perform a titration of $\ce{CH3COOH}$ with $\ce{NaOH}$. Let the original concentration of acetic acid be $\pu{0.2 M}$. The volumes of $\ce{CH3COOH}$ and $\ce{NaOH}$ are the same. Then the equivalence point will be reached when equivolume of $\pu{0.2 M}$ of $\ce{NaOH}$ has been added to the solution. At the equivalence point, the $\mathrm{pH}$ will be above 7. How does one calculate the exact $\mathrm{pH}$ value of the solution, which will be at the equilibrium point? The reactions we have are:

$$\ce{CH3COOH + OH- <=> CH3COO- + H2O} \tag1$$ ($\ce{Na^+}$ is a bystander or spectator ion, and therefore I ignored here).

The equilibrium constant for the given reaction is $K_1 = 1/K_b(\ce{CH3COO^-}) \gg 0$. We also have the following reaction:

For the reaction, $$\ce{CH3COO- + H3O+ <=> CH3COOH + H2O} \tag2$$, the equilibrium constant is $K_2 = 1/K_a(\ce{CH3COOH}) \gg 0$.

It is argued on other posts here that the first reaction "essentially" goes to completion and because $\ce{CH3COO-}$ later reacts with $\ce{H3O+}$ in the reaction $(2)$, the $\mathrm{pH}$ will be above 7. However, this motivation seems flawed to me. When the second reaction occurs, that will decrease the concentration of $\ce{CH3COO-}$, which will trigger reaction $(1)$ to move to the right. This will decrease the concentration of $\ce{OH^-$ ions. I know that the rate at which the concentration of $OH-}$ decreases in reaction $(1)$ and the rate at which the concentration of $\ce{H3O+}$ decreases in reaction $(2)$ are not equal due to the different equilibrium constants for these two reactions. However, I don't understand how to calculate this $\mathrm{pH}$ using the above logic.

Could anyone show how to calculate the $\mathrm{pH}$ that this titration will have at the equivalence point using logic above? Thanks!

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  • $\begingroup$ You have not specified what volumes of each reagent are being mixed together. $\endgroup$
    – Sam202
    Feb 19, 2023 at 17:46
  • $\begingroup$ Sorry, I have updated the question now. $\endgroup$
    – UserE
    Feb 19, 2023 at 17:48
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    $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Feb 19, 2023 at 17:55
  • $\begingroup$ Why have you omitted the reaction of acatate ion with water, being a weak base? $\endgroup$
    – Poutnik
    Feb 19, 2023 at 17:59
  • $\begingroup$ @Poutnik that is incorporated into the first reaction, right? $\endgroup$
    – UserE
    Feb 19, 2023 at 18:05

2 Answers 2

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At the equivalence point, there is excess neither of acetic acid neither of sodium hydroxide. The solution is equivalent to the solution of sodium acetate of the concentration $\pu{0.1 M}$, as both acetate and sodium ions are in the doubled volume now.

For all evaluation below, I assume the temperature $T_\mathrm{C} = \pu{25 ^\circ C}$. Everywhere is also silently assumed the ion activity coefficients are equal to 1.

We know that an alkaline salt of a weak acid with the acidity constant $K_\mathrm{a}$ acts as a weak base with the basicity constant $K_\mathrm{b} = K_\mathrm{w}/K_\mathrm{a}$ (and $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$):

$$K_b = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}= \frac{[\ce{HA}]K_\mathrm{w}}{[\ce{H+}][\ce{A-}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}} \tag{1}$$

For diluted solutions of weak acids, there is the approximating equation:

$$\mathrm{pH} \approx 0.5(\mathrm{p}K_\mathrm{a} - \log{c}) \label{2} \tag{2},$$

assuming conditions:

$$c \gg [\ce{H+}] \gg [\ce{OH-}]\tag{2a}.$$

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \overset{[\ce{H+}] \gg [\ce{OH-}]}\approx \frac{[\ce{H+}][\ce{H+}]}{[\ce{HA}]} \overset{c \gg [\ce{H+}]}\approx \frac{[\ce{H+}][\ce{H+}]}{c} \\ \implies [\ce{H+}]=\sqrt{K_\mathrm{a} \cdot c} \implies \log{([\ce{H+}])}=0.5(\log{(K_\mathrm{a})} + \log{(c)}) \implies \eqref{2} \tag{2b}$$

Similarly, for weak bases, there is the approximation: $$\mathrm{pOH} \approx 0.5(\mathrm{p}K_\mathrm{b} - \log{c})\tag{3},$$

assuming conditions:

$$c \gg [\ce{OH-}] \gg [\ce{H+}] \label{3a} \tag{3a}.$$

From it there is derived the approximated equation for $\mathrm{pH}$ of solution of a salt of a weak acid and a strong base at given molar concentration:

$$\mathrm{pH} = 14 - \mathrm{pOH} \approx 14 - 0.5(14 - \mathrm{p}K_\mathrm{a} - \log{c}) = 7 + 0.5(\mathrm{p}K_\mathrm{a} + \log{c}) \tag{4},$$

assuming the conditions \eqref{3a} are met.

For the particular case in the question:

$$\mathrm{pH} \approx 7 + 0.5(4.75 + \log{0.1}) \approx 8.88 \tag{5}.$$

Checking $\eqref{3a}$: $$\pu{0.1 M} \gg \pu{10^{-5.12} M} \gg \pu{10^{-8.88} M} \tag{6}$$

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  • $\begingroup$ Hmm, I have intentionally used t for Celsius temperature in contrary of T for absolute temperature. I have also intentionally used A/B instead of frac for easier reading and text flow. Similarly, I have not intentionally used MJ on standalone inline pH and pOH, but that can be subject of various opinions. And, no access seems better than niether access to me. $\endgroup$
    – Poutnik
    Feb 19, 2023 at 22:02
  • $\begingroup$ You can always roll back edits, and keep it like that or edit more yourself. I think on this site, the philosophy is a bit different than on Wikipedia ("no editor owns an article – any contributions can and may be mercilessly edited and redistributed"). Here, if you have a substantially different answer, you just post your own. However, we edit questions extensively when warranted, and we do sometimes edit answers as well. $\endgroup$
    – Karsten
    Feb 19, 2023 at 22:13
  • $\begingroup$ @Karsten I know and I did partial reediting of this. Personally, I hesitate to change intentions of the posts of others. I fix obvious errors or apply formatting that is out of the poster knowledge scope, but otherwise rather suggest changes. $\endgroup$
    – Poutnik
    Feb 19, 2023 at 22:17
  • $\begingroup$ @Poutnik what is equation 2 for theorem? I haven't seen it before. I tried googling it but found little. What is it called? $\endgroup$
    – UserE
    Feb 20, 2023 at 12:22
  • $\begingroup$ @UserE It is not a theorem and I am not sure if it has a particular name. It is equation for calculation of pH of diluted weak acid, simplified under specific conditions (2a), allowing simplification (2b). It is regularly taught in lectures about pH of water solutions of weak acids, bases and salts. $\endgroup$
    – Poutnik
    Feb 20, 2023 at 12:30
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If the expected pH is alkaline, the hydroxide concentration will be higher than the hydronium concentration. Thus, working with K$_1$ is easier (you can neglect the hydronium concentration and focus on the major species acetate and the minor species acetic acid and hydroxide). As the last step in the calculation, you can estimate the hydronium concentration from the hydroxide concentration.

The logic given by the OP does not follow a simple path to the solution, so I would not use it. Instead, you can first let reaction 1 go to completion, and then realize you overshot and go reverse a bit. That way, it looks like a straightforward weak base calculation.

Here are the concentrations of species after you add the two solutions, pretending that acetic acid does not partially dissociates, and that it does not react with sodium hydroxide:

$$c_\mathrm{acetic\ acid} = \pu{0.1 mol L-1}$$ $$c_\mathrm{acetate} = \pu{0.0 mol L-1}$$ $$c_\mathrm{hydroxide} = \pu{0.1 mol L-1}$$

The latter ignores water auto-dissociation.

Now we pretend the reaction goes to completion:

$$c_\mathrm{acetic\ acid} = \pu{0.0 mol L-1}$$ $$c_\mathrm{acetate} = \pu{0.1 mol L-1}$$ $$c_\mathrm{hydroxide} = \pu{0.0 mol L-1}$$

Again, we are ignoring the auto-dissociation of water.

This is not at equilibrium because acetate is a weak base. You can calculate the hydroxide and acetic acid concentrations (still ignoring the auto-dissociation of water) using the usual method you learned for weak bases. Then, you can calculate the concentration of hydronium ions, neglecting that when hydronium ions form, hydroxide ions also form (you can do that because the hydroxide concentration is much higher than the hydronium concentration, so the error is very small).

Once you have a set of approximate equilibrium concentrations, you can plug them into the equilibrium constant expression to verify that your approximations made sense (i.e. the equilibrium constants calculated are close to those known or given).

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  • $\begingroup$ Do you mean that you only consider reaction Nr. 1, and based on the equilibrium constant for that reaction, you calculate the concentration of [OH^-] at equilibrium? Then you estimate $[H_3O^+]$ as $[H_3O^+]=K_w/[OH^-]$. $\endgroup$
    – UserE
    Feb 19, 2023 at 18:16
  • $\begingroup$ @UserE Yes, exactly. $\endgroup$
    – Karsten
    Feb 19, 2023 at 19:16
  • $\begingroup$ Why is $c_{acetic\ acid}=0.1$ in your first equation? Should not it be 0.2 M? $\endgroup$
    – UserE
    Feb 20, 2023 at 12:46
  • $\begingroup$ When you mix it with the other solution (any solution, really), it gets diluted. $\endgroup$
    – Karsten
    Feb 20, 2023 at 13:00

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