In the derivation of the pH of a solution of a salt prepared from a weak acid and strong base, I used two equilibrium reactions:

$$\ce{ CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)} \qquad k_{eq} = K_a \\ \ce{ 2H2O(l) <=> H3O+(aq) + OH-(aq) } \qquad k_{eq} = K_w$$

Reversing the first equation and adding it to the second gave me the "hydrolysis reaction": $$ \ce{ CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq)} \qquad k_{eq} = \frac{K_w}{K_a} $$ My textbook uses this third equation to derive the pH of the salt solution by asserting that the equilibrium concentrations of the reactants and products of the third reaction are: $$ \ce{[CH3COO-(aq)]} = C-C\alpha \\ \ce{[CH3COOH(aq)]} = C\alpha \\ \ce{[OH-(aq)]} = C\alpha $$

What I am confused about is, why is the equilibrium concentration of hydroxide ions $C\alpha$, and is only governed by the third reaction? Why does the second equilibrium not contribute any hydroxide ion concentration? That is to say, the derivation assumes the water equilibrium does not exist, and all the hydroxide ion concentration comes only from the third reaction.

Also, even if we suppose some $x$ concentration of hydroxide ions and use it in the 'equilibrium constant expressions' of both the second and third reactions, we have more unknowns then conditions. So that's a problem too.

All in all, I need to know that, why, inspite of hydroxide ion participating in two equilibria, is the hydroxide ion concentration decided by only one of the reactions. And even if approximation is involved, if we try to be accurate, why do we have fewer conditions than variables?

Thank you.


1 Answer 1


In a solution of acetic ions, say from sodium acetate, some acetate will react with water to produce acetic acid and hydroxide anion (your "hydrolysis" equation"):

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

Now, let's say you dissolved 1.00 moles of sodium acetate in 1.00 L of water. For the sake of convenience, you would describe this solution as $\mathrm{1.00\ M\ }\ce{CH3COONa}$. However the actual concentration of acetate ion is less than $\mathrm{1.00\ M}$ because of the reaction with water.

Enter $C$, the "total concentration of acetate". This concentration is the sum of all concentrations of species that are related to or derived from acetate. In this case the relevant species are the acetate ion and acetic acid, so $C = [\ce{CH3COO-}]+[\ce{CH3COOH}]$.

What is $\alpha$? $\alpha$ represents that percent dissociation or percent ionization of the acetate ion. Therefore, $C\alpha$ represents the concentration of acetate that has ionization (reacted with water). In this case, since every acetate that reacts with water produces acetic acid, $C\alpha = [\ce{CH3COOH}]$.

How does this derivation lead to $C\alpha = [\ce{OH-}]$? You need to look back at the reaction equation for the process. For each acetate ion that reacts with water, one acetic acid and one hydroxide ion are produced. Based on stoichiometry, the concentrations of acetatic acid and hydroxide ion must be the same.

How do we determine $[\ce{OH-}]$? We plug the three expressions for the three species into the law of mass action and solve for $\alpha$ (we know $C$):

$$K=\dfrac{K_w}{K_a} = \dfrac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}=\dfrac{(C\alpha)^2}{C-C\alpha}$$

Now, if there was some additional hydroxide $x$ already present, say because the pH was fixed at 12.00, how do we handle that? The expressions for the concentrations of acetate and acetic acid are still valid. $C$ is still the total concentration of acetate + acetic acid. $C\alpha$ is still the concentration of acetic acid. $C-C\alpha$ is still the concentration of acetate ion. However, the concentration of hydroxide ion is different in this case. There is some concentration from the initial conditions $x$, and there is some increase from the ionization of acetate ion $C\alpha$.

If the initial hydorixde concentration is from the autoionization of water, we can account for that:

$$\begin{aligned} [\ce{CH3COO-}] &= C-C\alpha \\ [\ce{CH3COOH}] &= C\alpha \\ [\ce{OH-}] &= 1\times 10^{-7} + C\alpha \end{aligned}$$

However, except for the most dilute solutions, the contribution from the autoionization of water is usually orders of magtitude smaller than $C\alpha$ and gets lost in the insignificant figures. Let's compare the same calculation for our $1\ \mathrm{M}$ acetate solution. Note that you get the same outcome regardless if you include the hydroxide concentration form $K_w$.

$$\begin{array}{|c|c|c|} \hline & \text{without autoionization} & \text{with autoionization}\\ \hline C & \mathrm{1.00\ M} & \mathrm{1.00\ M} \\ [\ce{CH3COO-}] & C-C\alpha = 1-\alpha & C-C\alpha = 1-\alpha \\ [\ce{CH3COOH}] & C\alpha = \alpha & C\alpha = \alpha \\ [\ce{OH-}] & C\alpha = \alpha & 10^{-7} + C\alpha = 10^{-7} + \alpha \\ K & \dfrac{K_w}{K_a} = \dfrac{10^{-14}}{1.74\times 10^{-5}} & \dfrac{10^{-14}}{1.74\times 10^{-5}}\\ K & \dfrac{\alpha^2}{1-\alpha} & \dfrac{(10^{-7} + \alpha)(\alpha)}{1-\alpha}\\ \alpha & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\ [\ce{CH3COO-}] & 1.00-2.40 \times 10^{-5} \approx 1 & 1.00-2.39 \times 10^{-5} \approx 1 \\ [\ce{CH3COOH}] & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\ [\ce{OH-}] & 2.40 \times 10^{-5} & 10^{-7} + 2.39 \times 10^{-5} = 2.40 \times 10^{-5} \\ \hline \end{array}$$

  • $\begingroup$ Thank you for your answer! I understood it, and though helpful, I actually meant to ask something slightly different. For example: $\ce{A<=>C+D}$ and $\ce{B<=>C+F}$. Here C is formed from both, the 1st and 2nd reaction. So the final conc. of C has contribution from 2 reactions. Similarly, $\ce{OH-}$ should have contribution from both reactions, the 2nd and the 3rd. But we assume contribution from only 3rd reaction. Why? $\endgroup$ Jan 3, 2017 at 12:26
  • $\begingroup$ Well now it seems clear, just one more thing that bugged me: for the autoionization calculation, a contribution of $10^{-7}\,$M was assumed. But how can we assume that? I mean, why $10^{-7}$? Why not $10^{-6}$? Sure, neutral water has $10^{-7}$ conc. but that isn't the case here... wouldn't the contribution change or something? $\endgroup$ Jan 4, 2017 at 6:27
  • $\begingroup$ Perhaps, can we visualize adding the salt after the water equilibrium is attained? That would make sense. Is this correct? $\endgroup$ Jan 4, 2017 at 6:28

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