Calculate the pH at the equilibrium point in an acetic acid sodium hydroxide titration

Question

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!

Answer 1

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

Answer 2

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).