Understanding how to calculate pH of a buffer solution

[zenterix]

Homework Statement

You are working in a microbiology lab culturing bacteria that require an acidic environment and you need to prepare a buffer to keep the culture at an appropriate pH.

Relevant Equations

You prepare a buffer solution that is 0.040M $\mathrm{NaCH_3CO_2(aq)}$ and 0.080M $\mathrm{CH_3COOH(aq)}$ at $\mathrm{25^\circ C}$.

What is the pH of the solution?

I'd like to go through the steps in this calculation carefully to make sure I fully understand what is happening.

We start with a total of three chemical compounds: $\mathrm{H_2O, CH_3COOH}$ and $\mathrm{NaCH_3CO_2}$ at $\mathrm{25^\circ C}$.

That is, we have water, a weak acid, and a salt of the conjugate base of the weak acid.

When the weak acid is added to water, some of its molecules are deprotonated by water molecules and in the resulting solution we have $\mathrm{H_3O^+}$ and $\mathrm{CH_3CO_2^-}$.

When the salt of the conjugate base is added to water, it dissociates into ions $\mathrm{Na^+}$ and $\mathrm{CH_3CO_2^-}$.

The complete ionic equation seems to be

$$\mathrm{Na^+(aq)+CH_3CO_2^-(aq)+CH_3COOH(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+2CH_3CO_2^-(aq)+Na^+(aq)$$

and the net ionic equation is thus

$$\mathrm{CH_3COOH(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+CH_3CO_2^-(aq)}$$

Thus, the only reaction that is happening is that between a weak acid molecule and a water molecule, and between hydronium and the conjugate base.

These compounds form a dynamic equilibrium with equilibrium constant

$$K_a=\mathrm{\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3COOH]}}=1.8\times 10^{-5}}$$

We start with 0.080M of weak acid, 0M of hydronium, and 0.040M of the conjugate base.

xM of the weak acid deprotonates.

The equation for $K_a$ can be solved for $x$ which is $\mathrm{[H_3O^+]}$.

The result is a pH of 4.44.

Now, a few questions and observations

1) The salt of the conjugate base contributes an initial non-zero concentration of the conjugate base to the dynamic equilibrium.

2) $\mathrm{Na^+}$ does nothing. It is a spectator.

3) One fundamental doubt I have is about the relationship between strength of an acid and its conjugate base.

What I learned is that

$$K_a\times K_b=K_w$$

which leads to

$$pK_a+pK_b=pK_w=14.00\ \ \ \text{at}\ 25^\circ C$$

Consider the equations

$$\mathrm{NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)}$$

$$\mathrm{NH_4^+(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+NH_3(aq)}$$

These are the proton transfer equilibria for ammonia and its conjugate acid.

Experimentally, we have $pK_b=4.75$ and $pK_a=9.25$.

What I don't get is the following.

It seems that a lot of $\mathrm{NH_4^+}$ are deprotonated and not that many $\mathrm{NH_3}$ are protonated.

I guess this means that if we start with just ammonia, then we don't get a whole lot of ammonium, but whatever we do get of ammonium is rapidly protonated back to ammonia.

I guess at some point the rate of deprotonation of a large base of ammonia equals the protonation of a small base of ammonium and we get an equilibrium, which is characterized by the concentrations of ammonia and ammonium at which this equality of protonation and deprotonation occurs.

 

[Borek]

Basic idea behind buffer calculations is that once you add a weak acid and its salt into the solution they don't react (they do to some extent, but it is negligible). So you can safely assume 0.040 M and 0.080 M are the real concentrations of things present in the solution.

 

[zenterix]

If we were to use an unbuffered solution of just acetic acid, the equilibrium pH would be 2.92.

That is, we end up with a lot of hydronium (compared to the buffer).

In the buffered solution of the OP, however, we already start with a significant concentration of conjugate base, acetate.

The amount of hydronium in equilibrium is smaller than in the unbuffered solution.

In both cases, however, when we add some extra base, it appears that the approximation you are referring to can be used (ie, the approximation that acid and conjugate base concentrations don't change very much in the resulting reaction between the added base and the original weak acid, all that really changes is hydronium concentration).