How to calculate pH at equivalence point using this alternative?

[zenterix]

Homework Statement

Consider the following simple titration problem.

We have 50mL of 0.10M solution of acetic acid (aq).

We use as titrant a 0.20M solution of NaOH(aq).

Relevant Equations

The equivalence point is reached after adding 25mL of titrant.

That is, all of the initial 0.005 mol of acetic acid (present in the 50mL analyte) is neutralized by the 0.005 mol of NaOH (present in 25mL of titrant).

0.005 mol of $\mathrm{CH_3CO_2^-}$ is generated and the concentration of this conjugate base is

$$\mathrm{[CH_3COO_2^-]_{init}=\frac{1}{15}M}$$

where "init" denotes simply the concentration right after neutralization before equilibrium is reached.

Suppose we want to find the pH at the equivalence point.

The way I know how to do it is the following.

Acetic acid has a dynamic equilibrium with water in aqueous solution given by

$$\mathrm{CH_3COOH(aq)+H_2O(l)\rightleftharpoons CH_3COO_2^-(aq)+H_3O^+(aq)}\tag{1}$$

We are given $K_a=1.74\times 10^{-5}$.

The conjugate base of acetic acid is $\mathrm{CH_3CO_2^-}$ and it has a dynamic equilibrium with water in aqueous solution given by

$$\mathrm{CH_3CO_2^-(aq)+H_2O(l)\rightleftharpoons CH_3COOH(aq)+OH^-(aq)}\tag{2}$$

$$K_b=\frac{K_w}{K_a}=5.75\times 10^{-10}\tag{3}$$

Then

$$\mathrm{K_b=\frac{[CH_3COOH][OH^-]}{[CH_3CO_2^-]}=\frac{x^2}{\frac{1}{15}-x}}$$

$$\implies x=\mathrm{[OH^-]}=6.19\times 10^{-6}$$

and we get a pH of 8.79, which makes sense since $\mathrm{CH_3CO_2^-}$ is a weak base in water and at the equivalence point we have a solution of this weak base.

My question is the following.

Can we obtain the same result using (1) and $K_a$?

I am not sure how to do it.

In (2) we are saying that acetate does proton transfer with a water molecule generating acetic acid and hydroxide and we work with the corresponding equilibrium and equilibrium constant $K_b$.

In (1) we are saying that acetic acid does proton transfer with a water molecule generating acetate and hydronium. There is an equilibrium and an equilibrium constant $K_a$ that is related to $K_b$.

So, at the equivalence point, we start with 1/15 M of acetate, no acetic acid, and some very small amount of hydronium. The exact amount of hydronium can be computed by computing the pH of the initial solution with just acetic acid in equilibrium

$$\mathrm{1.74\times 10^{-5}=\frac{x^2}{0.10-x}\implies x=[H_3O^+]=1.3104\times 10^{-3}M}$$

and then computing the concentration after adding titrant to reach the equivalence point

$$\mathrm{\frac{1.3104\times 10^{-3}\cdot 0.050}{0.075}=0.0008736M}$$

But what to do next?

 

[zenterix]

Initially I thought to do

$$\mathrm{K_a=1.74\times 10^{-5}=\frac{[CH_3CO_2^-][H_3O^+]}{[CH_3COOH]}=\frac{(1/15-x)(0.0008736-x)}{x}}$$

 

[Borek]

Can we obtain the same result using (1) and $K_a$?

Not without using also K_w (which makes the process elaborate and convoluted).

 

[zenterix]

Ok. But how do we make use of $K_w$?

I am trying to understand this calculation as a way of understanding what is going on, even if it is more convoluted than necessary.

Where and how do I use $K_w$?

 

[Borek]

As usual: write formula for Ka, Kw, mass balances, charge balance, solve. This approach works always.

Simplified approaches have limited applicability (so have to be carefully selected to fit the problem), but when they work calculations are much easier and much faster.

 

[zenterix]

$$\mathrm{CH_3COOH(aq)+H_2O(l)\rightleftharpoons CH_3CO_2^-(aq)+H_3O^+(aq)}\tag{1}$$

$$\mathrm{K_a=1.74\times 10^{-5}=\frac{[CH_3CO_2^-][H_3O^+]}{[CH_3COOH]}}\tag{2}$$

$$\mathrm{K_w=[H_3O^+][OH^-]}\tag{3}$$

$$\mathrm{[H_3O^+]=[OH^-]+[CH_3CO_2^-]}\tag{4}$$

$$\mathrm{[CH_3CO_2^-]_{init}=[CH_3CO_2^-]+[CH_3COOH]}\tag{5}$$

We have the unknowns $\mathrm{[H_3O^+],[OH^-],[CH_3CO_2^-],}$ and $\mathrm{[CH_3COOH]}$ and we have four equations.

When I solve for $\mathrm{[H_3O^+]}$ I reach a value that is too high, so the pH is too low.

These equations are probably incorrect.

 

[Borek]

You added Na⁺ to the solution, so you should include it in the charge balance.

Also: I told you the "init" index is misleading. "Total" or "analytical" better describe what you mean. Plus using ions in place of total concentration (even with the index "init") is a simple way of getting lost at some point. Call it C_Tor C_A and you won't confuse symbols.

 

[zenterix]

Okay, I solved the problem, and indeed the issue is with the charge balance equation.

In my other thread where we discuss this I have to be honest: I was not convinced by your arguments that "total" or "analytical" are better subscripts.

"Analytical" seems to be alright in the sense that it describes a concentration as being theoretical rather than an actual concentration that occurs.

As in my other thread, this is how I am currently thinking about the charge balance equation

At any point in time, the ions in the solution cancel out in terms of charge

$$\mathrm{[Na^+]+[H_3O^+]=[OH^-]+[CH_3CO_2^-]}\tag{1}$$

If we put all terms on the lhs we get an equation that says that the sum of concentrations of positive ions minus the concentrations of negative ions is zero.

$$\mathrm{[Na^+]+[H_3O^+]-[OH^-]-[CH_3CO_2^-]=0}\tag{2}$$

And this is true always.

Recall that we are at the stoichiometric point after having added the required amount of titrant (a base).

The analyte (an acid) has been completely neutralized and this has generated a concentration of $1/15$M of its conjugate base, $\mathrm{CH_3CO_2^-}$.

This is what I am calling $\mathrm{[CH_3CO_2^-]_{init}}$.

You would like to replace "init" with "total" or "analytical. Sure, we can debate that and it is important for understanding the concepts but not for the calculation.

"init" emphasizes the fact that we are considering two points in time: the point in time right after neutralization but no adjustment and another later point in time at which equilibrium is reached.

Anyway, (2) gives us the equations

$$\mathrm{[Na^+]_{init}+[H_3O^+]_{init}-[OH^-]_{init}-[CH_3CO_2^-]_{init}=0}\tag{3}$$

$$\mathrm{[Na^+]_{eq}+[H_3O^+]_{eq}-[OH^-]_{eq}-[CH_3CO_2^-]_{eq}=0}\tag{4}$$

In theory, we have very small concentrations of hydronium and hydroxide in (3) but we can approximate them as zero because the concentration of acetate $\mathrm{CH_3CO_2^-]_{init}}$ is so large (and thus dominates the amount of charge at the "init" point in time).

We also have the equation

$$\mathrm{[Na^+]_{init}=[Na^+]_{eq}}\tag{5}$$

Thus, if we equate (3) and (4) we get (I am omitting the "eq" subscript, but this is implicit).

$$\mathrm{[CH_3CO_2^-]_{init}=[CH_3CO_2^-]+[OH^-]-[H_3O^+]}\tag{6}$$

We now have four equations and four unknowns and after some algebra we reach an expression for $K_a$ that is function of $\mathrm{[H_3O^+]}$ which we can solve and we reach the correct and expected value for pH of 8.79.

Note that (6) is the charge balance equation at equilibrium.

From what I understood, you think that writing one of

$$\mathrm{[CH_3CO_2^-]_{total}=[CH_3CO_2^-]+[OH^-]-[H_3O^+]}\tag{7a}$$

$$\mathrm{[CH_3CO_2^-]_{analytical}=[CH_3CO_2^-]+[OH^-]-[H_3O^+]}\tag{7b}$$

is less misleading.

I don't see how these are conceptually better or more correct than "init", especially not "total", but I am always open to new arguments or reformulations of previous arguments.

Also, it's not like "init" is my invention. It's the subscript used in the book by Atkins that I am reading.