How to take into account autoprotolysis in weak base solution?

[zenterix]

Homework Statement

Suppose we want to titrate a solution of formic acid, $\mathrm{HCOOH}$.

Relevant Equations

Suppose our initial goal is to compute the pH of the titrated solution at the stoichiometric point.

We start with 25ml of a 0.100M solution of $\mathrm{HCOOH(aq)}$ and as titrant we use a 0.150M solution of $\mathrm{NaOH(aq)}$.

For reference, we can calculate the initial pH of the formic acid solution, which comes out to 2.37.

I will go over just the results of the initial calculations, which I have no issues with.

In the sample of formic acid, there is 0.0025 moles of acid.

To neutralize all the acid we need 17ml of titrant (which contains 0.0025 moles of sodium hydroxide).

After neutralizing all the acid, we are left with a 0.06M solution of the salt of the conjugate base of formic acid, namely, $\mathrm{NaHCO_2^-}$.

This solution is basic.

We can calculate the pH in the following way.

The proton transfer reaction for the conjugate base is

$$\mathrm{HCO_2^-(aq)+H_2O(l)\rightleftharpoons HCO_2^-(aq)+OH^-(aq)}$$

with

$$K_b=5.6\times 10^{-11}=\frac{x^2}{0.06-x}$$

where $x=\mathrm{[OH^-]=[HCOOH]}$.

We solve for $x$ and obtain

$$x=\mathrm{[OH^-]=1.8\times 10^{-6}}$$

and so the pH of the salt solution at equilibrium is

$$\text{pH}=8.26$$

So now we get to the part I have trouble with.

Note that we have a solution of a weak base with an initial concentration of base that is 0.06M which is way above $10^{-6}$M, a threshold used for ignoring ions due to autoprotolysis of water.

Suppose we were to redo the calculation of the pH of the salt solution taking into account autoprotolysis of water.

My question is how exactly to do it.

Here is what I tried to do.

We have two equilibria, that of the weak base and that of autoprotolysis of water.

In the final solution we have the compounds $\mathrm{H_3O^+,OH^-,HCO_2^-,}$ and $\mathrm{HCOOH}$, besides some spectator ions.

We want the concentrations of these compounds and so we need four equations.

Two are the equilibrium constants

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

$$K_b=\mathrm{\frac{[HCOOH][OH^-]}{[HCO_2^-]}}$$

The other two equations are the charge balance and material balance equations.

I have doubts about how to write these.

Material balance seems to be

$$\mathrm{[HCO_2^-]_{init}=[HCO_2^-]+[HCOOH]}$$

Charge balance, however, I am very unsure of. I initially thought of

$$\mathrm{[HCO_2^-]+[OH^-]=[H_3O^+]}$$

Using these four equations I reach the following equation

$$\mathrm{K_b=\frac{([HCO_2^-]_{init}-\left ( \frac{K_w}{[OH^-]}-[OH^-] \right )[OH^-]}{\frac{K_w}{[OH^-]}-[OH^-]}}$$

Inputting known values and solving for $\mathrm{[OH^-]}$ gives us

$$\mathrm{[OH^-]=3.14\times 10^{-12}}$$

for a pOH of 11.50 and a pH of 2.49.

 

[zenterix]

I think I figured out the problem.

It is indeed the charge balance equation.

The solution has an initial charge due to $\mathrm{[HCO_2^-]_{init}}$.

Thus, charge balance requires

$$\mathrm{[HCO_2^-]_{init}=[HCO_2^-]+[OH^-]-[H_3O^+]}$$

Then, after putting all four equations together we get

$$\mathrm{K_b=\frac{\left ([HCO_2^-]_{init}-\left ( [HCO_2^-]_{init}-[OH^-]+\frac{K_w}{[OH^-]}\right )\right )[OH^-]}{[HCO_2^-]_{init}-[OH^-]+\frac{K_w}{[OH^-]}}}$$

and now solving for $\mathrm{[OH^-]}$ gives us

$$1.83\times 10^{-6}\text{M}$$

which is very close to the one obtained initially, ignoring autoprotolysis.

 

[Borek]

Charge balance doesn't contain "initial", it contains only real, actual values.

Accidentally what you did gives a correct answer as "initial" concentration of HCOO^- is equal to the Na⁺ that is present in the solution (and which you ignored in your charge balance equation).

Compare https://www.chembuddy.com/pH-calculation-lectures where you will find all the cases you asked about (and then some).

 

[Mayhem]

OP, you would do yourself a favor using ICE tables for these problems rather than manually writing everything out.

 

[zenterix]

Charge balance doesn't contain "initial", it contains only real, actual values.

Not sure what this means.

Two states are considered for the solution of the salt of conjugate base: a well-defined initial state and a final equilibrium state we want to find.

"initial" is a subscript on concentration of the conjugate base indicating the concentration is for the initial state.

$\mathrm{Na^+}$ is a spectator ion: its concentration is the same in both states.

In this problem, the conjugate base is composed of charged molecules.

The initial charge of the solution is due to $\mathrm{[HCO_2^-]_{initial}}$ and $\mathrm{[Na^+]}$ and the charge of the solution at equilibrium is due to $\mathrm{[H_3O+],[OH^-],[HCO_2^-],}$ and $\mathrm{[Na^+]}$.

$\mathrm{[Na^+]}$ appears on both sides so gets cancelled.

I was not clear about this in my last post when I said the "initial charge is due to $\mathrm{[HCO_2^-]_{initial}}$", which is an incomplete statement and so incorrect.

 

[zenterix]

OP, you would do yourself a favor using ICE tables for these problems rather than manually writing everything out.

Hadn't read about the term "ICE tables" but from a google search this seems to be just the technique I am already using.

In the OP, I wrote "I will go over just the results of the initial calculations, which I have no issues with.". These initial calculations include the use of an ICE table to calculate the desired pH.

It's just that I wanted to go through a more accurate calculation that includes autoprotolysis. Can't use an ICE table for this calculation, right?

After all, this is the point of the question in the OP.

 

[Borek]

$\mathrm{[Na^+]}$ appears on both sides so gets cancelled.

I think you are in some way mistaken about what charge balance is.

Charge balance for your problem is

[Na⁺] + [H⁺] = [OH^-] + [HCOO^-]

and these are all concentrations at equilibrium. Doesn't matter that the Na⁺ concentration doesn't change, it is still in the solution and still needs to be taken into account if the solution is to remain electrically neutral. It doesn't "cancel out" in any way.

"Initial" concentration of Na⁺ (after dissolution of HCOONa, but before any reactions kick in, nonexisting state, serves just as reference) is for obvious reasons identical to the initial concentration of HCOO^-.

 

[zenterix]

The equation you wrote can also be written as

$$\mathrm{[Na^+]+[H_3O^+]-[OH^-]-[HCO_2^-]=0}\tag{1}$$

which seems says that the total charge of the solution is always zero, no matter the point in time at which we take the concentrations of the solutes in the solution.

That is, it is true at equilibrium but also at the start of the reaction.

From what I understand, this equation expressses "charge balance": the charges in the solution always balance and result in zero charge.

At the start of the reaction, we have

$$\mathrm{[Na^+]_{init}+[H_3O^+]_{init}-[OH^-]_{init}-[HCO_2^-]_{init}=0}\tag{2}$$

But $\mathrm{[H_3O^+]_{init}=[OH^-]_{init}=0}$. Thus

$$\mathrm{[Na^+]_{init}-[HCO_2^-]_{init}=0}\tag{3}$$

At equilibrium we have

$$\mathrm{[Na^+]+[H_3O^+]-[OH^-]-[HCO_2^-]=0}\tag{4}$$

where each concentration denotes the equilibrium value.

We can equate (3) and (4) and use the extra equation $\mathrm{[Na^+]=const}$.

The resulting equation expresses that total charge in the solution is zero both initially and at equilibrium.

In this equation we mathematically cancel out the terms $\mathrm{[Na^+]}$ and $\mathrm{[Na^+]_{init}}$, resulting in the equation

$$\mathrm{[HCO_2^-]_{init}=[HCO_2^-]+[OH^-]-[H_3O^+]}\tag{5}$$

which is the "charge balance" equation I used in post #2.