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So the problem is this: is the sodium acetate acidic, basic or neutral?

One of my textbooks (ATAR Notes chemistry) says in solution, acetate simply (due to its negative charge) accepts protons and produces hydroxide ions

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-} \tag{1}$$

Hence the [$\ce{OH-}$] increases so solution is basic. Hence acetate is a basic anion.

However, my teacher's notes say:

\begin{align} \ce{CH3COO- + H3O+ &<=> CH3COOH + H2O (l)} \tag{2}\\ \ce{H3O+ + OH- &<=> H2O + H2O} \tag{3} \end{align}

that the acetate ions will be released into solution and thus, to decrease their concentration, some of the $\ce{CH3COO-}$ ions react with $\ce{H3O+}$ (LCP). Consequently, $\ce{H3O+}$ drops and more water ionizes and thus [$\ce{OH-}$] increases (LCP). So the acetate ion is a basic anion.

Which pathway, that leads to the formation of basic solution due to the acetate ion, is the correct one?

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  • $\begingroup$ Avoid using zeros instead of letters O in chemical formulas. It hints carelessness or ignorance and none of them is taken lightly. // You are not the first asking about this topic. It seems you did not do searching properly. $\endgroup$
    – Poutnik
    Commented Apr 25 at 6:51
  • $\begingroup$ Consider formatting guides for texts and formulas/equations/expressions $\endgroup$
    – Poutnik
    Commented Apr 25 at 6:54
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    $\begingroup$ Both approaches by Percival are correct. They are really similar. The second is a rewritten variation of the first pathway, using other words. $\endgroup$
    – Maurice
    Commented Apr 25 at 7:35
  • $\begingroup$ Both pathways are correct and simultaneous. Their equilibrium constants are related. As the reactions are written, the constants are Kb, 1/Ka, 1/Kw, where Ka . Kb = Kw = 10^(-14) at 25 deg C.. $\endgroup$
    – Poutnik
    Commented Apr 25 at 7:53
  • $\begingroup$ @poutnik would you care to share the link where this topic was asked elsewhere? I simply cannot find it. $\endgroup$ Commented Apr 26 at 1:46

2 Answers 2

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All reactions that you have provided are correct.

These reactions have the general schema, following Broensted-Lawry concept of acids and bases:

$$\ce{acid1 + base2 <=>[H+ exchange] base1 + acid2}$$ or $$\ce{HA + B <=> A- + BH+}$$

when the abstract $\ce{A}$ and $\ce{B}$ components may be both neutral or charged.

The acid may be in your case any particle providing $\ce{H+}$ ion, like $\ce{CH3COOH}$, $\ce{H3O+}$ or even (very weak one) $\ce{H2O}$.

The base may be any particle accepting $\ce{H+}$ ion: $\ce{CH3COO-}$, $\ce{OH-}$ or even (very weak one) $\ce{H2O}$.

Each of combinations has its own equilibrium constant, that can be derived from respective dissociation constants of acids or bases.

$\ce{CH3COO-}$ as the base reacts simultaneously with two acids:

  • the weak acid $\ce{H2O}$ that is in huge excess.
    • $\ce{CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq)}$
  • the strong acid $\ce{H3O+}$ that is sparse.
    • $\ce{CH3COO-(aq) + H3O+(aq) <=> CH3COOH(aq) + H2O(l)}$
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The $K_a$ of $\ce{CH3COOH}$ at room temperature is about:

$$K_a=1.8\cdot10^{-5}$$

That means the $K_b$ for the acetate ion is:

$$K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{1.8\cdot10^{-5}}=5.56\cdot10^{-10}$$

Now we can analyze the equilibrium system that forms when adding pure acetate into water. Assuming we initially start with $0.1M$ (i.e. a high enough concentration for water auto-ionization to be reasonably negligible):

$$\ce{CH3COO-(aq) +H2O(l)<=>CH3COOH(aq) +OH-(aq)}$$

At equilibrium, we have the following concentrations:

$$C_{\ce{CH3COO-}}=0.1-x$$

$$C_{\ce{CH3COOH}}=x$$

$$C_{\ce{OH-}}=x$$

Plugging them into the $K_b$ formula:

$$K_b=5.56\cdot10^{-10}=\frac{x^2}{0.1-x}$$

Solving for $x$:

$$x=\pu{7.41\cdot10^{-6}M}$$

So the final concentration of hydroxide would be:

$$C_{\ce{OH-}}=\pu{7.41\cdot10^{-6}M}$$

And the pH of this slightly basic solution would be approximately:

$$\pmb{\pu{pH}=14+\log(7.41\cdot10^{-6})=8.87}$$

If you were to repeat the analysis with $\frac{1}{10^4}$ fraction for the initial concentration of acetate, you would have to include the auto-ionization of water, and the resulting pH of this practically neutral solution would be approximately:

$$\pmb{\pu{pH}=14+\log(3.96\cdot10^{-8}+10^{-7})=7.14}$$

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  • $\begingroup$ Sam202's text is OK. But it is not an answer to the question. $\endgroup$
    – Maurice
    Commented Apr 25 at 7:32

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